Solve 4x^2-39=10

$$4 x ^ { 2 } - 39 = 10$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

Steps Using Difference of Squares

Subtract 10 from both sides.

$$4x^{2}-39-10=0$$

Subtract 10 from -39 to get -49.

$$4x^{2}-49=0$$

Consider 4x^{2}-49. Rewrite 4x^{2}-49 as \left(2x\right)^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2x-7\right)\left(2x+7\right)=0$$

To find equation solutions, solve 2x-7=0 and 2x+7=0.

$$x=\frac{7}{2} x=-\frac{7}{2}$$

Steps by Finding Square Root

Add 39 to both sides.

$$4x^{2}=10+39$$

Add 10 and 39 to get 49.

$$4x^{2}=49$$

Divide both sides by 4.

$$x^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{7}{2} x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Subtract 10 from both sides.

$$4x^{2}-39-10=0$$

Subtract 10 from -39 to get -49.

$$4x^{2}-49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-49\right)}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4\left(-49\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\left(-49\right)}}{2\times 4}$$

Multiply -16 times -49.

$$x=\frac{0±\sqrt{784}}{2\times 4}$$

Take the square root of 784.

$$x=\frac{0±28}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±28}{8}$$

Now solve the equation x=\frac{0±28}{8} when ± is plus. Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{0±28}{8} when ± is minus. Reduce the fraction \frac{-28}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{7}{2}$$

The equation is now solved.

$$x=\frac{7}{2} x=-\frac{7}{2}$$