3a^2+40a+50=0

Solution for 3a^2+40a+50=0 equation:

3a^2+40a+50=0

a = 3; b = 40; c = +50;
Δ = b2-4ac
Δ = 402-4·3·50
Δ = 1000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1000}=\sqrt{100*10}=\sqrt{100}*\sqrt{10}=10\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{10}}{2*3}=\frac{-40-10\sqrt{10}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{10}}{2*3}=\frac{-40+10\sqrt{10}}{6}
$