20x^2+56x+15=0

Solution for 20x^2+56x+15=0 equation:

20x^2+56x+15=0

a = 20; b = 56; c = +15;
Δ = b2-4ac
Δ = 562-4·20·15
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-44}{2*20}=\frac{-100}{40}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+44}{2*20}=\frac{-12}{40}
=-3/10
$