Solve (x-5)*(x+5)=75

$$( x - 5 ) \cdot ( x + 5 ) = 75$$

Solve for x

$$x=-10$$

$$x=10$$

Steps by Finding Square Root

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-25=75$$

Add 25 to both sides.

$$x^{2}=75+25$$

Add 75 and 25 to get 100.

$$x^{2}=100$$

Take the square root of both sides of the equation.

$$x=10 x=-10$$

Steps Using the Quadratic Formula

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-25=75$$

Subtract 75 from both sides.

$$x^{2}-25-75=0$$

Subtract 75 from -25 to get -100.

$$x^{2}-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-100\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-100\right)}}{2}$$

Multiply -4 times -100.

$$x=\frac{0±\sqrt{400}}{2}$$

Take the square root of 400.

$$x=\frac{0±20}{2}$$

Now solve the equation x=\frac{0±20}{2} when ± is plus. Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{0±20}{2} when ± is minus. Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=10 x=-10$$