(x)(3x+2)=270

Solution for (x)(3x+2)=270 equation:

(x)(3x+2)=270

We move all terms to the left:

(x)(3x+2)-(270)=0

We multiply parentheses

3x^2+2x-270=0

a = 3; b = 2; c = -270;
Δ = b2-4ac
Δ = 22-4·3·(-270)
Δ = 3244
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:

$\sqrt{\Delta}=\sqrt{3244}=\sqrt{4*811}=\sqrt{4}*\sqrt{811}=2\sqrt{811}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{811}}{2*3}=\frac{-2-2\sqrt{811}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{811}}{2*3}=\frac{-2+2\sqrt{811}}{6}
$