(3-4)2=9x2-+16

Solution for (3-4)2=9x2-+16 equation:

(3-4)2=9x^2-+16

We move all terms to the left:

(3-4)2-(9x^2-+16)=0

We add all the numbers together, and all the variables

-(9x^2-+16)+(-1)2=0

We add all the numbers together, and all the variables

-(9x^2-+16)-2=0

We use the square of the difference formula

-(9x^2-16)-2=0

We get rid of parentheses

-9x^2+16-2=0

We add all the numbers together, and all the variables

-9x^2+14=0

a = -9; b = 0; c = +14;
Δ = b2-4ac
Δ = 02-4·(-9)·14
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*-9}=\frac{0-6\sqrt{14}}{-18}
=-\frac{6\sqrt{14}}{-18}
=-\frac{\sqrt{14}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*-9}=\frac{0+6\sqrt{14}}{-18}
=\frac{6\sqrt{14}}{-18}
=\frac{\sqrt{14}}{-3}
$