2x^2+4x-48=598

Solution for 2x^2+4x-48=598 equation:

2x^2+4x-48=598

We move all terms to the left:

2x^2+4x-48-(598)=0

We add all the numbers together, and all the variables

2x^2+4x-646=0

a = 2; b = 4; c = -646;
Δ = b2-4ac
Δ = 42-4·2·(-646)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-72}{2*2}=\frac{-76}{4}
=-19
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+72}{2*2}=\frac{68}{4}
=17
$