Solve 1.25x^2+15x=2.5x^2+40

$$1.25 { x }^{ 2 } +15x=2.5 { x }^{ 2 } +40$$

Solve for x

$$x=4$$

$$x=8$$

Steps Using the Quadratic Formula

Subtract 2.5x^{2} from both sides.

$$1.25x^{2}+15x-2.5x^{2}=40$$

Combine 1.25x^{2} and -2.5x^{2} to get -1.25x^{2}.

$$-1.25x^{2}+15x=40$$

Subtract 40 from both sides.

$$-1.25x^{2}+15x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1.25 for a, 15 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\left(-1.25\right)\left(-40\right)}}{2\left(-1.25\right)}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\left(-1.25\right)\left(-40\right)}}{2\left(-1.25\right)}$$

Multiply -4 times -1.25.

$$x=\frac{-15±\sqrt{225+5\left(-40\right)}}{2\left(-1.25\right)}$$

Multiply 5 times -40.

$$x=\frac{-15±\sqrt{225-200}}{2\left(-1.25\right)}$$

Add 225 to -200.

$$x=\frac{-15±\sqrt{25}}{2\left(-1.25\right)}$$

Take the square root of 25.

$$x=\frac{-15±5}{2\left(-1.25\right)}$$

Multiply 2 times -1.25.

$$x=\frac{-15±5}{-2.5}$$

Now solve the equation x=\frac{-15±5}{-2.5} when ± is plus. Add -15 to 5.

$$x=-\frac{10}{-2.5}$$

Divide -10 by -2.5 by multiplying -10 by the reciprocal of -2.5.

$$x=4$$

Now solve the equation x=\frac{-15±5}{-2.5} when ± is minus. Subtract 5 from -15.

$$x=-\frac{20}{-2.5}$$

Divide -20 by -2.5 by multiplying -20 by the reciprocal of -2.5.

$$x=8$$

The equation is now solved.

$$x=4 x=8$$

Steps for Completing the Square

Subtract 2.5x^{2} from both sides.

$$1.25x^{2}+15x-2.5x^{2}=40$$

Combine 1.25x^{2} and -2.5x^{2} to get -1.25x^{2}.

$$-1.25x^{2}+15x=40$$

Divide both sides of the equation by -1.25, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-1.25x^{2}+15x}{-1.25}=\frac{40}{-1.25}$$

Dividing by -1.25 undoes the multiplication by -1.25.

$$x^{2}+\frac{15}{-1.25}x=\frac{40}{-1.25}$$

Divide 15 by -1.25 by multiplying 15 by the reciprocal of -1.25.

$$x^{2}-12x=\frac{40}{-1.25}$$

Divide 40 by -1.25 by multiplying 40 by the reciprocal of -1.25.

$$x^{2}-12x=-32$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-32+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-32+36$$

Add -32 to 36.

$$x^{2}-12x+36=4$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-6=2 x-6=-2$$

Add 6 to both sides of the equation.

$$x=8 x=4$$