Solve {x^2+7x/2-57}=0

$$\{ x ^ { 2 } + \frac { 7 x } { 2 } - 57 \} = 0$$

Solve for x

$$x = -\frac{19}{2} = -9\frac{1}{2} = -9.5$$

$$x=6$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$2x^{2}+7x-114=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-114. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\left(-114\right)=-228$$

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 -228.

$$-1,228 -2,114 -3,76 -4,57 -6,38 -12,19$$

Calculate the sum for each pair.

$$-1+228=227 -2+114=112 -3+76=73 -4+57=53 -6+38=32 -12+19=7$$

The solution is the pair that gives sum 7.

$$a=-12 b=19$$

Rewrite 2x^{2}+7x-114 as \left(2x^{2}-12x\right)+\left(19x-114\right).

$$\left(2x^{2}-12x\right)+\left(19x-114\right)$$

Factor out 2x in the first and 19 in the second group.

$$2x\left(x-6\right)+19\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(2x+19\right)$$

To find equation solutions, solve x-6=0 and 2x+19=0.

$$x=6 x=-\frac{19}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2x^{2}+7x-114=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -114 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\left(-114\right)}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\left(-114\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\left(-114\right)}}{2\times 2}$$

Multiply -8 times -114.

$$x=\frac{-7±\sqrt{49+912}}{2\times 2}$$

Add 49 to 912.

$$x=\frac{-7±\sqrt{961}}{2\times 2}$$

Take the square root of 961.

$$x=\frac{-7±31}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±31}{4}$$

Now solve the equation x=\frac{-7±31}{4} when ± is plus. Add -7 to 31.

$$x=\frac{24}{4}$$

Divide 24 by 4.

$$x=6$$

Now solve the equation x=\frac{-7±31}{4} when ± is minus. Subtract 31 from -7.

$$x=-\frac{38}{4}$$

Reduce the fraction \frac{-38}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{19}{2}$$

The equation is now solved.

$$x=6 x=-\frac{19}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2x^{2}+7x-114=0$$

Add 114 to both sides. Anything plus zero gives itself.

$$2x^{2}+7x=114$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=\frac{114}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=\frac{114}{2}$$

Divide 114 by 2.

$$x^{2}+\frac{7}{2}x=57$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=57+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=57+\frac{49}{16}$$

Add 57 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{961}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{961}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{961}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{31}{4} x+\frac{7}{4}=-\frac{31}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=6 x=-\frac{19}{2}$$