Solve -28x+5=36x^2

$$-28x+5 = 36 { x }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{94}-7}{18}\approx 0.149742206$$

$$x=\frac{-\sqrt{94}-7}{18}\approx -0.927519984$$

Steps Using the Quadratic Formula

Subtract 36x^{2} from both sides.

$$-28x+5-36x^{2}=0$$

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.

$$-36x^{2}-28x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -36 for a, -28 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\left(-36\right)\times 5}}{2\left(-36\right)}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\left(-36\right)\times 5}}{2\left(-36\right)}$$

Multiply -4 times -36.

$$x=\frac{-\left(-28\right)±\sqrt{784+144\times 5}}{2\left(-36\right)}$$

Multiply 144 times 5.

$$x=\frac{-\left(-28\right)±\sqrt{784+720}}{2\left(-36\right)}$$

Add 784 to 720.

$$x=\frac{-\left(-28\right)±\sqrt{1504}}{2\left(-36\right)}$$

Take the square root of 1504.

$$x=\frac{-\left(-28\right)±4\sqrt{94}}{2\left(-36\right)}$$

The opposite of -28 is 28.

$$x=\frac{28±4\sqrt{94}}{2\left(-36\right)}$$

Multiply 2 times -36.

$$x=\frac{28±4\sqrt{94}}{-72}$$

Now solve the equation x=\frac{28±4\sqrt{94}}{-72} when ± is plus. Add 28 to 4\sqrt{94}.

$$x=\frac{4\sqrt{94}+28}{-72}$$

Divide 28+4\sqrt{94} by -72.

$$x=\frac{-\sqrt{94}-7}{18}$$

Now solve the equation x=\frac{28±4\sqrt{94}}{-72} when ± is minus. Subtract 4\sqrt{94} from 28.

$$x=\frac{28-4\sqrt{94}}{-72}$$

Divide 28-4\sqrt{94} by -72.

$$x=\frac{\sqrt{94}-7}{18}$$

The equation is now solved.

$$x=\frac{-\sqrt{94}-7}{18} x=\frac{\sqrt{94}-7}{18}$$

Steps for Completing the Square

Subtract 36x^{2} from both sides.

$$-28x+5-36x^{2}=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-28x-36x^{2}=-5$$

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.

$$-36x^{2}-28x=-5$$

Divide both sides by -36.

$$\frac{-36x^{2}-28x}{-36}=-\frac{5}{-36}$$

Dividing by -36 undoes the multiplication by -36.

$$x^{2}+\left(-\frac{28}{-36}\right)x=-\frac{5}{-36}$$

Reduce the fraction \frac{-28}{-36} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{7}{9}x=-\frac{5}{-36}$$

Divide -5 by -36.

$$x^{2}+\frac{7}{9}x=\frac{5}{36}$$

Divide \frac{7}{9}, the coefficient of the x term, by 2 to get \frac{7}{18}. Then add the square of \frac{7}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{9}x+\left(\frac{7}{18}\right)^{2}=\frac{5}{36}+\left(\frac{7}{18}\right)^{2}$$

Square \frac{7}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{9}x+\frac{49}{324}=\frac{5}{36}+\frac{49}{324}$$

Add \frac{5}{36} to \frac{49}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{9}x+\frac{49}{324}=\frac{47}{162}$$

Factor x^{2}+\frac{7}{9}x+\frac{49}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{18}\right)^{2}=\frac{47}{162}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{18}\right)^{2}}=\sqrt{\frac{47}{162}}$$

Simplify.

$$x+\frac{7}{18}=\frac{\sqrt{94}}{18} x+\frac{7}{18}=-\frac{\sqrt{94}}{18}$$

Subtract \frac{7}{18} from both sides of the equation.

$$x=\frac{\sqrt{94}-7}{18} x=\frac{-\sqrt{94}-7}{18}$$