Solve 2x^2+2x-5=x^2-6x+4

$$2 x ^ { 2 } + 2 x - 5 = x ^ { 2 } - 6 x + 4$$

Solve for x

$$x=-9$$

$$x=1$$

Steps Using Factoring

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=-6x+4$$

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

$$x^{2}+2x-5=-6x+4$$

Add 6x to both sides.

$$x^{2}+2x-5+6x=4$$

Combine 2x and 6x to get 8x.

$$x^{2}+8x-5=4$$

Subtract 4 from both sides.

$$x^{2}+8x-5-4=0$$

Subtract 4 from -5 to get -9.

$$x^{2}+8x-9=0$$

To solve the equation, factor x^{2}+8x-9 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.

$$a+b=8 ab=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

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

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=-1 b=9$$

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

$$\left(x-1\right)\left(x+9\right)$$

To find equation solutions, solve x-1=0 and x+9=0.

$$x=1 x=-9$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=-6x+4$$

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

$$x^{2}+2x-5=-6x+4$$

Add 6x to both sides.

$$x^{2}+2x-5+6x=4$$

Combine 2x and 6x to get 8x.

$$x^{2}+8x-5=4$$

Subtract 4 from both sides.

$$x^{2}+8x-5-4=0$$

Subtract 4 from -5 to get -9.

$$x^{2}+8x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-9\right)=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

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

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=-1 b=9$$

Rewrite x^{2}+8x-9 as \left(x^{2}-x\right)+\left(9x-9\right).

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

Factor out x in the first and 9 in the second group.

$$x\left(x-1\right)+9\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+9\right)$$

To find equation solutions, solve x-1=0 and x+9=0.

$$x=1 x=-9$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=-6x+4$$

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

$$x^{2}+2x-5=-6x+4$$

Add 6x to both sides.

$$x^{2}+2x-5+6x=4$$

Combine 2x and 6x to get 8x.

$$x^{2}+8x-5=4$$

Subtract 4 from both sides.

$$x^{2}+8x-5-4=0$$

Subtract 4 from -5 to get -9.

$$x^{2}+8x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-9\right)}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-8±\sqrt{64+36}}{2}$$

Add 64 to 36.

$$x=\frac{-8±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-8±10}{2}$$

Now solve the equation x=\frac{-8±10}{2} when ± is plus. Add -8 to 10.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-8±10}{2} when ± is minus. Subtract 10 from -8.

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

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=1 x=-9$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=-6x+4$$

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

$$x^{2}+2x-5=-6x+4$$

Add 6x to both sides.

$$x^{2}+2x-5+6x=4$$

Combine 2x and 6x to get 8x.

$$x^{2}+8x-5=4$$

Add 5 to both sides.

$$x^{2}+8x=4+5$$

Add 4 and 5 to get 9.

$$x^{2}+8x=9$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=9+4^{2}$$

Square 4.

$$x^{2}+8x+16=9+16$$

Add 9 to 16.

$$x^{2}+8x+16=25$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+4=5 x+4=-5$$

Subtract 4 from both sides of the equation.

$$x=1 x=-9$$