3x^2+34x-40=0

Solution for 3x^2+34x-40=0 equation:

3x^2+34x-40=0

a = 3; b = 34; c = -40;
Δ = b2-4ac
Δ = 342-4·3·(-40)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{409}}{2*3}=\frac{-34-2\sqrt{409}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{409}}{2*3}=\frac{-34+2\sqrt{409}}{6}
$