3x^2+4x^2+9=160

Solution for 3x^2+4x^2+9=160 equation:

3x^2+4x^2+9=160

We move all terms to the left:

3x^2+4x^2+9-(160)=0

We add all the numbers together, and all the variables

7x^2-151=0

a = 7; b = 0; c = -151;
Δ = b2-4ac
Δ = 02-4·7·(-151)
Δ = 4228
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:

$\sqrt{\Delta}=\sqrt{4228}=\sqrt{4*1057}=\sqrt{4}*\sqrt{1057}=2\sqrt{1057}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1057}}{2*7}=\frac{0-2\sqrt{1057}}{14}
=-\frac{2\sqrt{1057}}{14}
=-\frac{\sqrt{1057}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1057}}{2*7}=\frac{0+2\sqrt{1057}}{14}
=\frac{2\sqrt{1057}}{14}
=\frac{\sqrt{1057}}{7}
$