Solve 2x^2-5x-3=x^2+5x

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

Solve for x

$$x=2\sqrt{7}+5\approx 10.291502622$$

$$x=5-2\sqrt{7}\approx -0.291502622$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Subtract 5x from both sides.

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

Combine -5x and -5x to get -10x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -10.

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

Multiply -4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+12}}{2}$$

Add 100 to 12.

$$x=\frac{-\left(-10\right)±\sqrt{112}}{2}$$

Take the square root of 112.

$$x=\frac{-\left(-10\right)±4\sqrt{7}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±4\sqrt{7}}{2}$$

Now solve the equation x=\frac{10±4\sqrt{7}}{2} when ± is plus. Add 10 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}+10}{2}$$

Divide 10+4\sqrt{7} by 2.

$$x=2\sqrt{7}+5$$

Now solve the equation x=\frac{10±4\sqrt{7}}{2} when ± is minus. Subtract 4\sqrt{7} from 10.

$$x=\frac{10-4\sqrt{7}}{2}$$

Divide 10-4\sqrt{7} by 2.

$$x=5-2\sqrt{7}$$

The equation is now solved.

$$x=2\sqrt{7}+5 x=5-2\sqrt{7}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Subtract 5x from both sides.

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

Combine -5x and -5x to get -10x.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=3+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=3+25$$

Add 3 to 25.

$$x^{2}-10x+25=28$$

Factor x^{2}-10x+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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{28}$$

Simplify.

$$x-5=2\sqrt{7} x-5=-2\sqrt{7}$$

Add 5 to both sides of the equation.

$$x=2\sqrt{7}+5 x=5-2\sqrt{7}$$