Solve 10x=3+5x^2

$$10x=3+5 { x }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{10}}{5}+1\approx 1.632455532$$

$$x=-\frac{\sqrt{10}}{5}+1\approx 0.367544468$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

$$10x-3=5x^{2}$$

Subtract 5x^{2} from both sides.

$$10x-3-5x^{2}=0$$

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.

$$-5x^{2}+10x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-10±\sqrt{100+20\left(-3\right)}}{2\left(-5\right)}$$

Multiply 20 times -3.

$$x=\frac{-10±\sqrt{100-60}}{2\left(-5\right)}$$

Add 100 to -60.

$$x=\frac{-10±\sqrt{40}}{2\left(-5\right)}$$

Take the square root of 40.

$$x=\frac{-10±2\sqrt{10}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-10±2\sqrt{10}}{-10}$$

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

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

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

$$x=-\frac{\sqrt{10}}{5}+1$$

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

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

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

$$x=\frac{\sqrt{10}}{5}+1$$

The equation is now solved.

$$x=-\frac{\sqrt{10}}{5}+1 x=\frac{\sqrt{10}}{5}+1$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

$$10x-5x^{2}=3$$

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.

$$-5x^{2}+10x=3$$

Divide both sides by -5.

$$\frac{-5x^{2}+10x}{-5}=\frac{3}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{10}{-5}x=\frac{3}{-5}$$

Divide 10 by -5.

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

Divide 3 by -5.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-\frac{3}{5}+1$$

Add -\frac{3}{5} to 1.

$$x^{2}-2x+1=\frac{2}{5}$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=\frac{2}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{2}{5}}$$

Simplify.

$$x-1=\frac{\sqrt{10}}{5} x-1=-\frac{\sqrt{10}}{5}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{10}}{5}+1 x=-\frac{\sqrt{10}}{5}+1$$