Solve -5x^2+4x=035

$$- 5 x ^ { 2 } + 4 x = 035$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x=0$$

Steps Using Factoring

Multiply 0 and 35 to get 0.

$$-5x^{2}+4x=0$$

Factor out x.

$$x\left(-5x+4\right)=0$$

To find equation solutions, solve x=0 and -5x+4=0.

$$x=0 x=\frac{4}{5}$$

Steps Using the Quadratic Formula

Multiply 0 and 35 to get 0.

$$-5x^{2}+4x=0$$

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

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-4±4}{-10}$$

Now solve the equation x=\frac{-4±4}{-10} when ± is plus. Add -4 to 4.

$$x=\frac{0}{-10}$$

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-10} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{-10}$$

Reduce the fraction \frac{-8}{-10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=0 x=\frac{4}{5}$$

Steps for Completing the Square

Multiply 0 and 35 to get 0.

$$-5x^{2}+4x=0$$

Divide both sides by -5.

$$\frac{-5x^{2}+4x}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{4}{-5}x=\frac{0}{-5}$$

Divide 4 by -5.

$$x^{2}-\frac{4}{5}x=\frac{0}{-5}$$

Divide 0 by -5.

$$x^{2}-\frac{4}{5}x=0$$

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

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

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{4}{5} x=0$$