6 7 X 1 2

$\exponential{(x)}{2} - 6 10 - 7 $

\left(ten-7\correct)\left(10+1\correct)

\left(x-7\right)\left(10+1\correct)

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a+b=-6 ab=1\left(-vii\correct)=-7

Factor the expression by grouping. First, the expression needs to be rewritten every bit x^{2}+ax+bx-7. To find a and b, gear up a organization to exist solved.

a=-vii b=i

Since ab is negative, a and b have the contrary signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the organisation solution.

\left(10^{2}-7x\right)+\left(ten-7\right)

Rewrite x^{two}-6x-7 as \left(10^{2}-7x\right)+\left(ten-vii\right).

x\left(ten-seven\correct)+ten-seven

Factor out x in x^{2}-7x.

\left(x-seven\right)\left(10+ane\right)

Factor out common term ten-7 by using distributive holding.

ten^{2}-6x-7=0

Quadratic polynomial tin be factored using the transformation ax^{two}+bx+c=a\left(x-x_{one}\right)\left(10-x_{two}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

x=\frac{-\left(-6\correct)±\sqrt{\left(-6\right)^{2}-four\left(-7\right)}}{2}

All equations of the course ax^{two}+bx+c=0 tin can be solved using the quadratic formula: \frac{-b±\sqrt{b^{ii}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and ane when it is subtraction.

ten=\frac{-\left(-6\right)±\sqrt{36-iv\left(-7\correct)}}{two}

Square -half-dozen.

ten=\frac{-\left(-6\right)±\sqrt{36+28}}{2}

Multiply -4 times -7.

x=\frac{-\left(-6\right)±\sqrt{64}}{2}

Add 36 to 28.

x=\frac{-\left(-6\right)±eight}{two}

Take the foursquare root of 64.

x=\frac{half-dozen±viii}{2}

The reverse of -half dozen is half-dozen.

x=\frac{14}{ii}

Now solve the equation ten=\frac{half dozen±viii}{ii} when ± is plus. Add together six to eight.

10=\frac{-2}{2}

Now solve the equation x=\frac{6±8}{two} when ± is minus. Decrease 8 from 6.

x^{2}-6x-vii=\left(x-7\right)\left(x-\left(-1\right)\right)

Gene the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{ii}\right). Substitute vii for x_{1} and -i for x_{two}.

x^{2}-6x-7=\left(x-vii\right)\left(x+one\right)

Simplify all the expressions of the course p-\left(-q\correct) to p+q.

x ^ 2 -6x -7 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the straight factoring method, the equation must be in the form ten^2+Bx+C=0.

r + south = 6 rs = -7

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(ten−r)(x−s) where sum of factors (r+s)=−B and the production of factors rs = C

r = 3 - u due south = iii + u

2 numbers r and s sum upward to vi exactly when the average of the ii numbers is \frac{1}{ii}*6 = 3. You can also see that the midpoint of r and s corresponds to the centrality of symmetry of the parabola represented by the quadratic equation y=x^ii+Bx+C. The values of r and s are equidistant from the heart by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(3 - u) (3 + u) = -7

To solve for unknown quantity u, substitute these in the product equation rs = -7

ix - u^2 = -vii

Simplify past expanding (a -b) (a + b) = a^ii – b^2

-u^two = -7-nine = -16

Simplify the expression by subtracting 9 on both sides

u^two = 16 u = \pm\sqrt{16} = \pm four

Simplify the expression past multiplying -one on both sides and take the foursquare root to obtain the value of unknown variable u

r =3 - 4 = -1 s = 3 + four = vii

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

6 7 X 1 2,

Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20-6x-7

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