Solve x+3/2(2x+4+x-5)=60

$$\frac { x + 3 } { 2 } ( 2 x + 4 + x - 5 ) = 60$$

Solve for x

$$x = \frac{\sqrt{385} - 4}{3} \approx 5.207138957$$

$$x=\frac{-\sqrt{385}-4}{3}\approx -7.873805623$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$\left(x+3\right)\left(2x+4+x-5\right)=120$$

Combine 2x and x to get 3x.

$$\left(x+3\right)\left(3x+4-5\right)=120$$

Subtract 5 from 4 to get -1.

$$\left(x+3\right)\left(3x-1\right)=120$$

Apply the distributive property by multiplying each term of x+3 by each term of 3x-1.

$$3x^{2}-x+9x-3=120$$

Combine -x and 9x to get 8x.

$$3x^{2}+8x-3=120$$

Subtract 120 from both sides.

$$3x^{2}+8x-3-120=0$$

Subtract 120 from -3 to get -123.

$$3x^{2}+8x-123=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and -123 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\left(-123\right)}}{2\times 3}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 3\left(-123\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-8±\sqrt{64-12\left(-123\right)}}{2\times 3}$$

Multiply -12 times -123.

$$x=\frac{-8±\sqrt{64+1476}}{2\times 3}$$

Add 64 to 1476.

$$x=\frac{-8±\sqrt{1540}}{2\times 3}$$

Take the square root of 1540.

$$x=\frac{-8±2\sqrt{385}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-8±2\sqrt{385}}{6}$$

Now solve the equation x=\frac{-8±2\sqrt{385}}{6} when ± is plus. Add -8 to 2\sqrt{385}.

$$x=\frac{2\sqrt{385}-8}{6}$$

Divide -8+2\sqrt{385} by 6.

$$x=\frac{\sqrt{385}-4}{3}$$

Now solve the equation x=\frac{-8±2\sqrt{385}}{6} when ± is minus. Subtract 2\sqrt{385} from -8.

$$x=\frac{-2\sqrt{385}-8}{6}$$

Divide -8-2\sqrt{385} by 6.

$$x=\frac{-\sqrt{385}-4}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{385}-4}{3} x=\frac{-\sqrt{385}-4}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$\left(x+3\right)\left(2x+4+x-5\right)=120$$

Combine 2x and x to get 3x.

$$\left(x+3\right)\left(3x+4-5\right)=120$$

Subtract 5 from 4 to get -1.

$$\left(x+3\right)\left(3x-1\right)=120$$

Apply the distributive property by multiplying each term of x+3 by each term of 3x-1.

$$3x^{2}-x+9x-3=120$$

Combine -x and 9x to get 8x.

$$3x^{2}+8x-3=120$$

Add 3 to both sides.

$$3x^{2}+8x=120+3$$

Add 120 and 3 to get 123.

$$3x^{2}+8x=123$$

Divide both sides by 3.

$$\frac{3x^{2}+8x}{3}=\frac{123}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{8}{3}x=\frac{123}{3}$$

Divide 123 by 3.

$$x^{2}+\frac{8}{3}x=41$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=41+\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=41+\frac{16}{9}$$

Add 41 to \frac{16}{9}.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{385}{9}$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{385}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{385}{9}}$$

Simplify.

$$x+\frac{4}{3}=\frac{\sqrt{385}}{3} x+\frac{4}{3}=-\frac{\sqrt{385}}{3}$$

Subtract \frac{4}{3} from both sides of the equation.

$$x=\frac{\sqrt{385}-4}{3} x=\frac{-\sqrt{385}-4}{3}$$