3x^2+26x+55=0

Solution for 3x^2+26x+55=0 equation:

3x^2+26x+55=0

a = 3; b = 26; c = +55;
Δ = b2-4ac
Δ = 262-4·3·55
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4}{2*3}=\frac{-30}{6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4}{2*3}=\frac{-22}{6}
=-3+2/3
$