3a^2+10a-5=0

Solution for 3a^2+10a-5=0 equation:

3a^2+10a-5=0

a = 3; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·3·(-5)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{10}}{2*3}=\frac{-10-4\sqrt{10}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{10}}{2*3}=\frac{-10+4\sqrt{10}}{6}
$