Solve (3x^2-4)^2-3x^2=2(8+13x)

$$( 3 x ^ { 2 } - 4 ) ^ { 2 } - 3 x ^ { 2 } = 2 ( 8 + 13 x )$$

Solve for x

$$x = \frac{9 ^ {\frac{2}{3}} {(\sqrt[3]{2 \sqrt{22} + 13} + \sqrt[3]{13 - 2 \sqrt{22}})}}{9} \approx 2.092920182$$

$$x=0$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x^{2}-4\right)^{2}.

$$9\left(x^{2}\right)^{2}-24x^{2}+16-3x^{2}=2\left(8+13x\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$9x^{4}-24x^{2}+16-3x^{2}=2\left(8+13x\right)$$

Combine -24x^{2} and -3x^{2} to get -27x^{2}.

$$9x^{4}-27x^{2}+16=2\left(8+13x\right)$$

Use the distributive property to multiply 2 by 8+13x.

$$9x^{4}-27x^{2}+16=16+26x$$

Subtract 16 from both sides.

$$9x^{4}-27x^{2}+16-16=26x$$

Subtract 16 from 16 to get 0.

$$9x^{4}-27x^{2}=26x$$

Subtract 26x from both sides.

$$9x^{4}-27x^{2}-26x=0$$

Substitute t for x^{2}.

$$9t^{2}-27t-26=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 9 for a, -27 for b, and -26 for c in the quadratic formula.

$$t=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 9\left(-26\right)}}{2\times 9}$$

Do the calculations.

$$t=\frac{27±3\sqrt{185}}{18}$$

Solve the equation t=\frac{27±3\sqrt{185}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{185}}{6}+\frac{3}{2} t=-\frac{\sqrt{185}}{6}+\frac{3}{2}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=\frac{\sqrt{\frac{2\sqrt{185}}{3}+6}}{2} x=-\frac{\sqrt{\frac{2\sqrt{185}}{3}+6}}{2}$$