3x^2+20x+25=11

Solution for 3x^2+20x+25=11 equation:

3x^2+20x+25=11

We move all terms to the left:

3x^2+20x+25-(11)=0

We add all the numbers together, and all the variables

3x^2+20x+14=0

a = 3; b = 20; c = +14;
Δ = b2-4ac
Δ = 202-4·3·14
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{58}}{2*3}=\frac{-20-2\sqrt{58}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{58}}{2*3}=\frac{-20+2\sqrt{58}}{6}
$