8x^2+20x-28=0

Solution for 8x^2+20x-28=0 equation:

8x^2+20x-28=0

a = 8; b = 20; c = -28;
Δ = b2-4ac
Δ = 202-4·8·(-28)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-36}{2*8}=\frac{-56}{16}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+36}{2*8}=\frac{16}{16}
=1
$