Solve 3x^2+4=7

$$3 x ^ { 2 } + 4 = 7$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using Difference of Squares

Subtract 7 from both sides.

$$3x^{2}+4-7=0$$

Subtract 7 from 4 to get -3.

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

Divide both sides by 3.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

Steps by Finding Square Root

Subtract 4 from both sides.

$$3x^{2}=7-4$$

Subtract 4 from 7 to get 3.

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

Divide both sides by 3.

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

Divide 3 by 3 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Subtract 7 from both sides.

$$3x^{2}+4-7=0$$

Subtract 7 from 4 to get -3.

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

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

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

Square 0.

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

Multiply -4 times 3.

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

Multiply -12 times -3.

$$x=\frac{0±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$x=\frac{0±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±6}{6}$$

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

$$x=1$$

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

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$