8x^2+20x=48

Solution for 8x^2+20x=48 equation:

8x^2+20x=48

We move all terms to the left:

8x^2+20x-(48)=0

a = 8; b = 20; c = -48;
Δ = b2-4ac
Δ = 202-4·8·(-48)
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{-(20)-44}{2*8}=\frac{-64}{16}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+44}{2*8}=\frac{24}{16}
=1+1/2
$