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\left(4x-3\right)\left(4x+3\right)=0
Consider 16x^{2}-9. Rewrite 16x^{2}-9 as \left(4x\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
x=\frac{3}{4} x=-\frac{3}{4}
To find equation solutions, solve 4x-3=0 and 4x+3=0.
16x^{2}=9
Add 9 to both sides. Anything plus zero gives itself.
x^{2}=\frac{9}{16}
Divide both sides by 16.
x=\frac{3}{4} x=-\frac{3}{4}
Take the square root of both sides of the equation.
16x^{2}-9=0
Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.
x=\frac{0±\sqrt{0^{2}-4\times 16\left(-9\right)}}{2\times 16}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\times 16\left(-9\right)}}{2\times 16}
Square 0.
x=\frac{0±\sqrt{-64\left(-9\right)}}{2\times 16}
Multiply -4 times 16.
x=\frac{0±\sqrt{576}}{2\times 16}
Multiply -64 times -9.
x=\frac{0±24}{2\times 16}
Take the square root of 576.
x=\frac{0±24}{32}
Multiply 2 times 16.
x=\frac{3}{4}
Now solve the equation x=\frac{0±24}{32} when ± is plus. Reduce the fraction \frac{24}{32} to lowest terms by extracting and canceling out 8.
x=-\frac{3}{4}
Now solve the equation x=\frac{0±24}{32} when ± is minus. Reduce the fraction \frac{-24}{32} to lowest terms by extracting and canceling out 8.
x=\frac{3}{4} x=-\frac{3}{4}
The equation is now solved.