Solve x^2+6x-7=4x-x^2+8

$$x ^ { 2 } + 6 x - 7 = 4 x - x ^ { 2 } + 8$$

Solve for x

$$x = \frac{\sqrt{31} - 1}{2} \approx 2.283882181$$

$$x=\frac{-\sqrt{31}-1}{2}\approx -3.283882181$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$x^{2}+6x-7-4x=-x^{2}+8$$

Combine 6x and -4x to get 2x.

$$x^{2}+2x-7=-x^{2}+8$$

Add x^{2} to both sides.

$$x^{2}+2x-7+x^{2}=8$$

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

$$2x^{2}+2x-7=8$$

Subtract 8 from both sides.

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

Subtract 8 from -7 to get -15.

$$2x^{2}+2x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$x=\frac{-2±\sqrt{4+120}}{2\times 2}$$

Add 4 to 120.

$$x=\frac{-2±\sqrt{124}}{2\times 2}$$

Take the square root of 124.

$$x=\frac{-2±2\sqrt{31}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2\sqrt{31}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{31}}{4} when ± is plus. Add -2 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}-2}{4}$$

Divide -2+2\sqrt{31} by 4.

$$x=\frac{\sqrt{31}-1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{31}}{4} when ± is minus. Subtract 2\sqrt{31} from -2.

$$x=\frac{-2\sqrt{31}-2}{4}$$

Divide -2-2\sqrt{31} by 4.

$$x=\frac{-\sqrt{31}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{31}-1}{2} x=\frac{-\sqrt{31}-1}{2}$$

Steps for Completing the Square

Subtract 4x from both sides.

$$x^{2}+6x-7-4x=-x^{2}+8$$

Combine 6x and -4x to get 2x.

$$x^{2}+2x-7=-x^{2}+8$$

Add x^{2} to both sides.

$$x^{2}+2x-7+x^{2}=8$$

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

$$2x^{2}+2x-7=8$$

Add 7 to both sides.

$$2x^{2}+2x=8+7$$

Add 8 and 7 to get 15.

$$2x^{2}+2x=15$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{15}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{15}{2}$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{15}{2}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{15}{2}+\frac{1}{4}$$

Add \frac{15}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{31}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{31}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{31}}{2} x+\frac{1}{2}=-\frac{\sqrt{31}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{31}-1}{2} x=\frac{-\sqrt{31}-1}{2}$$