Solve (x^2+3)div2=14

$$( { x }^{ 2 } +3) \div 2=14$$

Solve for x

$$x=5$$

$$x=-5$$

Steps Using Difference of Squares

Multiply both sides by 2.

$$x^{2}+3=14\times 2$$

Multiply 14 and 2 to get 28.

$$x^{2}+3=28$$

Subtract 28 from both sides.

$$x^{2}+3-28=0$$

Subtract 28 from 3 to get -25.

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

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

Multiply both sides by 2.

$$x^{2}+3=14\times 2$$

Multiply 14 and 2 to get 28.

$$x^{2}+3=28$$

Subtract 3 from both sides.

$$x^{2}=28-3$$

Subtract 3 from 28 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

Multiply both sides by 2.

$$x^{2}+3=14\times 2$$

Multiply 14 and 2 to get 28.

$$x^{2}+3=28$$

Subtract 28 from both sides.

$$x^{2}+3-28=0$$

Subtract 28 from 3 to get -25.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$x=\frac{0±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{0±10}{2}$$

Now solve the equation x=\frac{0±10}{2} when ± is plus. Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{0±10}{2} when ± is minus. Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$