Solve 4a^2+2/2=3

$$\frac { 4 a ^ { 2 } + 2 } { 2 } = 3$$

Solve for a

$$a=1$$

$$a=-1$$

Steps Using Difference of Squares

Multiply both sides by 2.

$$4a^{2}+2=3\times 2$$

Multiply 3 and 2 to get 6.

$$4a^{2}+2=6$$

Subtract 6 from both sides.

$$4a^{2}+2-6=0$$

Subtract 6 from 2 to get -4.

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

Divide both sides by 4.

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

Consider a^{2}-1. Rewrite a^{2}-1 as a^{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).

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

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

$$a=1 a=-1$$

Steps by Finding Square Root

Multiply both sides by 2.

$$4a^{2}+2=3\times 2$$

Multiply 3 and 2 to get 6.

$$4a^{2}+2=6$$

Subtract 2 from both sides.

$$4a^{2}=6-2$$

Subtract 2 from 6 to get 4.

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

Divide both sides by 4.

$$a^{2}=\frac{4}{4}$$

Divide 4 by 4 to get 1.

$$a^{2}=1$$

Take the square root of both sides of the equation.

$$a=1 a=-1$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$4a^{2}+2=3\times 2$$

Multiply 3 and 2 to get 6.

$$4a^{2}+2=6$$

Subtract 6 from both sides.

$$4a^{2}+2-6=0$$

Subtract 6 from 2 to get -4.

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

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

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

Square 0.

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

Multiply -4 times 4.

$$a=\frac{0±\sqrt{-16\left(-4\right)}}{2\times 4}$$

Multiply -16 times -4.

$$a=\frac{0±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$a=\frac{0±8}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{0±8}{8}$$

Now solve the equation a=\frac{0±8}{8} when ± is plus. Divide 8 by 8.

$$a=1$$

Now solve the equation a=\frac{0±8}{8} when ± is minus. Divide -8 by 8.

$$a=-1$$

The equation is now solved.

$$a=1 a=-1$$