5x^2+16x-4=0

Solution for 5x^2+16x-4=0 equation:

5x^2+16x-4=0

a = 5; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·5·(-4)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{21}}{2*5}=\frac{-16-4\sqrt{21}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{21}}{2*5}=\frac{-16+4\sqrt{21}}{10}
$