2x2+10x=48

Solution for 2x2+10x=48 equation:

2x^2+10x=48

We move all terms to the left:

2x^2+10x-(48)=0

a = 2; b = 10; c = -48;
Δ = b2-4ac
Δ = 102-4·2·(-48)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*2}=\frac{12}{4}
=3
$