65x^2+8x-48=0

Solution for 65x^2+8x-48=0 equation:

65x^2+8x-48=0

a = 65; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·65·(-48)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-112}{2*65}=\frac{-120}{130}
=-12/13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+112}{2*65}=\frac{104}{130}
=4/5
$