Solve 2x(x-5)+3x=10(1/2-x)

$$2 x ( x - 5 ) + 3 x = 10 ( \frac { 1 } { 2 } - x )$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-5.

$$2x^{2}-10x+3x=10\left(\frac{1}{2}-x\right)$$

Combine -10x and 3x to get -7x.

$$2x^{2}-7x=10\left(\frac{1}{2}-x\right)$$

Use the distributive property to multiply 10 by \frac{1}{2}-x.

$$2x^{2}-7x=10\times \frac{1}{2}-10x$$

Multiply 10 and \frac{1}{2} to get \frac{10}{2}.

$$2x^{2}-7x=\frac{10}{2}-10x$$

Divide 10 by 2 to get 5.

$$2x^{2}-7x=5-10x$$

Subtract 5 from both sides.

$$2x^{2}-7x-5=-10x$$

Add 10x to both sides.

$$2x^{2}-7x-5+10x=0$$

Combine -7x and 10x to get 3x.

$$2x^{2}+3x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-3±\sqrt{9+40}}{2\times 2}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-3±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±7}{4}$$

Now solve the equation x=\frac{-3±7}{4} when ± is plus. Add -3 to 7.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-3±7}{4} when ± is minus. Subtract 7 from -3.

$$x=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=1 x=-\frac{5}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-5.

$$2x^{2}-10x+3x=10\left(\frac{1}{2}-x\right)$$

Combine -10x and 3x to get -7x.

$$2x^{2}-7x=10\left(\frac{1}{2}-x\right)$$

Use the distributive property to multiply 10 by \frac{1}{2}-x.

$$2x^{2}-7x=10\times \frac{1}{2}-10x$$

Multiply 10 and \frac{1}{2} to get \frac{10}{2}.

$$2x^{2}-7x=\frac{10}{2}-10x$$

Divide 10 by 2 to get 5.

$$2x^{2}-7x=5-10x$$

Add 10x to both sides.

$$2x^{2}-7x+10x=5$$

Combine -7x and 10x to get 3x.

$$2x^{2}+3x=5$$

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{5}{2}$$

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

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{5}{2}+\left(\frac{3}{4}\right)^{2}$$

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

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}$$

Add \frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{7}{4} x+\frac{3}{4}=-\frac{7}{4}$$

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

$$x=1 x=-\frac{5}{2}$$