Solve (x^2+10x-7575)=0

$$( { x }^{ 2 } +10x-7575)=0$$

Solve for x

$$x=20\sqrt{19}-5\approx 82.177978871$$

$$x=-20\sqrt{19}-5\approx -92.177978871$$

Steps Using the Quadratic Formula

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.

$$x^{2}+10x-7575=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -7575 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-7575\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-7575\right)}}{2}$$

Multiply -4 times -7575.

$$x=\frac{-10±\sqrt{100+30300}}{2}$$

Add 100 to 30300.

$$x=\frac{-10±\sqrt{30400}}{2}$$

Take the square root of 30400.

$$x=\frac{-10±40\sqrt{19}}{2}$$

Now solve the equation x=\frac{-10±40\sqrt{19}}{2} when ± is plus. Add -10 to 40\sqrt{19}.

$$x=\frac{40\sqrt{19}-10}{2}$$

Divide -10+40\sqrt{19} by 2.

$$x=20\sqrt{19}-5$$

Now solve the equation x=\frac{-10±40\sqrt{19}}{2} when ± is minus. Subtract 40\sqrt{19} from -10.

$$x=\frac{-40\sqrt{19}-10}{2}$$

Divide -10-40\sqrt{19} by 2.

$$x=-20\sqrt{19}-5$$

The equation is now solved.

$$x=20\sqrt{19}-5 x=-20\sqrt{19}-5$$

Steps for Completing the Square

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.

$$x^{2}+10x-7575=0$$

Add 7575 to both sides of the equation.

$$x^{2}+10x-7575-\left(-7575\right)=-\left(-7575\right)$$

Subtracting -7575 from itself leaves 0.

$$x^{2}+10x=-\left(-7575\right)$$

Subtract -7575 from 0.

$$x^{2}+10x=7575$$

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.

$$x^{2}+10x+5^{2}=7575+5^{2}$$

Square 5.

$$x^{2}+10x+25=7575+25$$

Add 7575 to 25.

$$x^{2}+10x+25=7600$$

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}.

$$\left(x+5\right)^{2}=7600$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{7600}$$

Simplify.

$$x+5=20\sqrt{19} x+5=-20\sqrt{19}$$

Subtract 5 from both sides of the equation.

$$x=20\sqrt{19}-5 x=-20\sqrt{19}-5$$