Solve -7x^2-12x+17=-10-8x^2

$$- 7 x ^ { 2 } - 12 x + 17 = - 10 - 8 x ^ { 2 }$$

Solve for x

$$x=3$$

$$x=9$$

Steps Using Factoring

Subtract -10 from both sides.

$$-7x^{2}-12x+17-\left(-10\right)=-8x^{2}$$

The opposite of -10 is 10.

$$-7x^{2}-12x+17+10=-8x^{2}$$

Add 8x^{2} to both sides.

$$-7x^{2}-12x+17+10+8x^{2}=0$$

Add 17 and 10 to get 27.

$$-7x^{2}-12x+27+8x^{2}=0$$

Combine -7x^{2} and 8x^{2} to get x^{2}.

$$x^{2}-12x+27=0$$

To solve the equation, factor x^{2}-12x+27 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-12 ab=27$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 27.

$$-1,-27 -3,-9$$

Calculate the sum for each pair.

$$-1-27=-28 -3-9=-12$$

The solution is the pair that gives sum -12.

$$a=-9 b=-3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-9\right)\left(x-3\right)$$

To find equation solutions, solve x-9=0 and x-3=0.

$$x=9 x=3$$

Steps Using Factoring By Grouping

Subtract -10 from both sides.

$$-7x^{2}-12x+17-\left(-10\right)=-8x^{2}$$

The opposite of -10 is 10.

$$-7x^{2}-12x+17+10=-8x^{2}$$

Add 8x^{2} to both sides.

$$-7x^{2}-12x+17+10+8x^{2}=0$$

Add 17 and 10 to get 27.

$$-7x^{2}-12x+27+8x^{2}=0$$

Combine -7x^{2} and 8x^{2} to get x^{2}.

$$x^{2}-12x+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 27=27$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 27.

$$-1,-27 -3,-9$$

Calculate the sum for each pair.

$$-1-27=-28 -3-9=-12$$

The solution is the pair that gives sum -12.

$$a=-9 b=-3$$

Rewrite x^{2}-12x+27 as \left(x^{2}-9x\right)+\left(-3x+27\right).

$$\left(x^{2}-9x\right)+\left(-3x+27\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-9\right)-3\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x-3\right)$$

To find equation solutions, solve x-9=0 and x-3=0.

$$x=9 x=3$$

Steps Using the Quadratic Formula

Subtract -10 from both sides.

$$-7x^{2}-12x+17-\left(-10\right)=-8x^{2}$$

The opposite of -10 is 10.

$$-7x^{2}-12x+17+10=-8x^{2}$$

Add 8x^{2} to both sides.

$$-7x^{2}-12x+17+10+8x^{2}=0$$

Add 17 and 10 to get 27.

$$-7x^{2}-12x+27+8x^{2}=0$$

Combine -7x^{2} and 8x^{2} to get x^{2}.

$$x^{2}-12x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 27}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 27}}{2}$$

Multiply -4 times 27.

$$x=\frac{-\left(-12\right)±\sqrt{144-108}}{2}$$

Add 144 to -108.

$$x=\frac{-\left(-12\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-12\right)±6}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±6}{2}$$

Now solve the equation x=\frac{12±6}{2} when ± is plus. Add 12 to 6.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{12±6}{2} when ± is minus. Subtract 6 from 12.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=9 x=3$$

Steps for Completing the Square

Add 8x^{2} to both sides.

$$-7x^{2}-12x+17+8x^{2}=-10$$

Combine -7x^{2} and 8x^{2} to get x^{2}.

$$x^{2}-12x+17=-10$$

Subtract 17 from both sides.

$$x^{2}-12x=-10-17$$

Subtract 17 from -10 to get -27.

$$x^{2}-12x=-27$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-27+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-27+36$$

Add -27 to 36.

$$x^{2}-12x+36=9$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-6=3 x-6=-3$$

Add 6 to both sides of the equation.

$$x=9 x=3$$