Solve 3-2x^2=10x

$$3 - 2 x ^ { 2 } = 10 x$$

Solve for x

$$x=\frac{\sqrt{31}-5}{2}\approx 0.283882181$$

$$x=\frac{-\sqrt{31}-5}{2}\approx -5.283882181$$

Steps Using the Quadratic Formula

Subtract 10x from both sides.

$$3-2x^{2}-10x=0$$

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.

$$-2x^{2}-10x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100+24}}{2\left(-2\right)}$$

Add 100 to 24.

$$x=\frac{-\left(-10\right)±\sqrt{124}}{2\left(-2\right)}$$

Take the square root of 124.

$$x=\frac{-\left(-10\right)±2\sqrt{31}}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{31}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±2\sqrt{31}}{-4}$$

Now solve the equation x=\frac{10±2\sqrt{31}}{-4} when ± is plus. Add 10 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}+10}{-4}$$

Divide 10+2\sqrt{31} by -4.

$$x=\frac{-\sqrt{31}-5}{2}$$

Now solve the equation x=\frac{10±2\sqrt{31}}{-4} when ± is minus. Subtract 2\sqrt{31} from 10.

$$x=\frac{10-2\sqrt{31}}{-4}$$

Divide 10-2\sqrt{31} by -4.

$$x=\frac{\sqrt{31}-5}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{31}-5}{2} x=\frac{\sqrt{31}-5}{2}$$

Steps for Completing the Square

Subtract 10x from both sides.

$$3-2x^{2}-10x=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-10x=-3$$

Divide both sides by -2.

$$\frac{-2x^{2}-10x}{-2}=-\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -10 by -2.

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

Divide -3 by -2.

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

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

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

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

$$x^{2}+5x+\frac{25}{4}=\frac{3}{2}+\frac{25}{4}$$

Add \frac{3}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=\frac{31}{4}$$

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

$$\left(x+\frac{5}{2}\right)^{2}=\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{31}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{31}}{2} x+\frac{5}{2}=-\frac{\sqrt{31}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{31}-5}{2} x=\frac{-\sqrt{31}-5}{2}$$