12x^2+51x-48=0

Solution for 12x^2+51x-48=0 equation:

12x^2+51x-48=0

a = 12; b = 51; c = -48;
Δ = b2-4ac
Δ = 512-4·12·(-48)
Δ = 4905
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4905}=\sqrt{9*545}=\sqrt{9}*\sqrt{545}=3\sqrt{545}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-3\sqrt{545}}{2*12}=\frac{-51-3\sqrt{545}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+3\sqrt{545}}{2*12}=\frac{-51+3\sqrt{545}}{24}
$