Solve a^2+3=7

$$a ^ { 2 } + 3 = 7$$

Solve for a

$$a=2$$

$$a=-2$$

Steps Using Difference of Squares

Subtract 7 from both sides.

$$a^{2}+3-7=0$$

Subtract 7 from 3 to get -4.

$$a^{2}-4=0$$

Consider a^{2}-4. Rewrite a^{2}-4 as a^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(a-2\right)\left(a+2\right)=0$$

To find equation solutions, solve a-2=0 and a+2=0.

$$a=2 a=-2$$

Steps by Finding Square Root

Subtract 3 from both sides.

$$a^{2}=7-3$$

Subtract 3 from 7 to get 4.

$$a^{2}=4$$

Take the square root of both sides of the equation.

$$a=2 a=-2$$

Steps Using the Quadratic Formula

Subtract 7 from both sides.

$$a^{2}+3-7=0$$

Subtract 7 from 3 to get -4.

$$a^{2}-4=0$$

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

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

Square 0.

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

Multiply -4 times -4.

$$a=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$a=\frac{0±4}{2}$$

Now solve the equation a=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$a=2$$

Now solve the equation a=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$a=-2$$

The equation is now solved.

$$a=2 a=-2$$