3x^2+26x+37=0

Solution for 3x^2+26x+37=0 equation:

3x^2+26x+37=0

a = 3; b = 26; c = +37;
Δ = b2-4ac
Δ = 262-4·3·37
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{-(26)-2\sqrt{58}}{2*3}=\frac{-26-2\sqrt{58}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{58}}{2*3}=\frac{-26+2\sqrt{58}}{6}
$