3x^2+38x-55=0

Solution for 3x^2+38x-55=0 equation:

3x^2+38x-55=0

a = 3; b = 38; c = -55;
Δ = b2-4ac
Δ = 382-4·3·(-55)
Δ = 2104
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:

$\sqrt{\Delta}=\sqrt{2104}=\sqrt{4*526}=\sqrt{4}*\sqrt{526}=2\sqrt{526}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{526}}{2*3}=\frac{-38-2\sqrt{526}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{526}}{2*3}=\frac{-38+2\sqrt{526}}{6}
$