Solve 16x^2+10=131

$$16 x ^ { 2 } + 10 = 131$$

Solve for x

$$x = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

$$x = -\frac{11}{4} = -2\frac{3}{4} = -2.75$$

Steps Using Difference of Squares

Subtract 131 from both sides.

$$16x^{2}+10-131=0$$

Subtract 131 from 10 to get -121.

$$16x^{2}-121=0$$

Consider 16x^{2}-121. Rewrite 16x^{2}-121 as \left(4x\right)^{2}-11^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4x-11\right)\left(4x+11\right)=0$$

To find equation solutions, solve 4x-11=0 and 4x+11=0.

$$x=\frac{11}{4} x=-\frac{11}{4}$$

Steps by Finding Square Root

Subtract 10 from both sides.

$$16x^{2}=131-10$$

Subtract 10 from 131 to get 121.

$$16x^{2}=121$$

Divide both sides by 16.

$$x^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$x=\frac{11}{4} x=-\frac{11}{4}$$

Steps Using the Quadratic Formula

Subtract 131 from both sides.

$$16x^{2}+10-131=0$$

Subtract 131 from 10 to get -121.

$$16x^{2}-121=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 0 for b, and -121 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 16\left(-121\right)}}{2\times 16}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 16\left(-121\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{0±\sqrt{-64\left(-121\right)}}{2\times 16}$$

Multiply -64 times -121.

$$x=\frac{0±\sqrt{7744}}{2\times 16}$$

Take the square root of 7744.

$$x=\frac{0±88}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{0±88}{32}$$

Now solve the equation x=\frac{0±88}{32} when ± is plus. Reduce the fraction \frac{88}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{11}{4}$$

Now solve the equation x=\frac{0±88}{32} when ± is minus. Reduce the fraction \frac{-88}{32} to lowest terms by extracting and canceling out 8.

$$x=-\frac{11}{4}$$

The equation is now solved.

$$x=\frac{11}{4} x=-\frac{11}{4}$$