Solve -7=5x-2x^2-7

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

Solve for x

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

$$x=0$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

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

Add 7 to both sides.

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

Add -7 and 7 to get 0.

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

Factor out x.

$$x\left(5-2x\right)=0$$

To find equation solutions, solve x=0 and 5-2x=0.

$$x=0 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

Add 7 to both sides.

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

Add -7 and 7 to get 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-5±5}{-4}$$

Now solve the equation x=\frac{-5±5}{-4} when ± is plus. Add -5 to 5.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-5±5}{-4} when ± is minus. Subtract 5 from -5.

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

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

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

The equation is now solved.

$$x=0 x=\frac{5}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Add 7 to both sides.

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

Add -7 and 7 to get 0.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -2.

$$\frac{-2x^{2}+5x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{5}{-2}x=\frac{0}{-2}$$

Divide 5 by -2.

$$x^{2}-\frac{5}{2}x=\frac{0}{-2}$$

Divide 0 by -2.

$$x^{2}-\frac{5}{2}x=0$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{5}{4} x-\frac{5}{4}=-\frac{5}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5}{2} x=0$$