Solve 3x^2+3=78

$$3 x ^ { 2 } + 3 = 78$$

Solve for x

$$x=5$$

$$x=-5$$

Steps Using Difference of Squares

Subtract 78 from both sides.

$$3x^{2}+3-78=0$$

Subtract 78 from 3 to get -75.

$$3x^{2}-75=0$$

Divide both sides by 3.

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

Consider x^{2}-25. Rewrite x^{2}-25 as x^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-5\right)\left(x+5\right)=0$$

To find equation solutions, solve x-5=0 and x+5=0.

$$x=5 x=-5$$

Steps by Finding Square Root

Subtract 3 from both sides.

$$3x^{2}=78-3$$

Subtract 3 from 78 to get 75.

$$3x^{2}=75$$

Divide both sides by 3.

$$x^{2}=\frac{75}{3}$$

Divide 75 by 3 to get 25.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

Steps Using the Quadratic Formula

Subtract 78 from both sides.

$$3x^{2}+3-78=0$$

Subtract 78 from 3 to get -75.

$$3x^{2}-75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-75\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-75\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-75\right)}}{2\times 3}$$

Multiply -12 times -75.

$$x=\frac{0±\sqrt{900}}{2\times 3}$$

Take the square root of 900.

$$x=\frac{0±30}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±30}{6}$$

Now solve the equation x=\frac{0±30}{6} when ± is plus. Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{0±30}{6} when ± is minus. Divide -30 by 6.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$