28a2+16a-80=0

Solution for 28a2+16a-80=0 equation:

28a^2+16a-80=0

a = 28; b = 16; c = -80;
Δ = b2-4ac
Δ = 162-4·28·(-80)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-96}{2*28}=\frac{-112}{56}
=-2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+96}{2*28}=\frac{80}{56}
=1+3/7
$