0=24x^2+71x-16

Solution for 0=24x^2+71x-16 equation:

0=24x^2+71x-16

We move all terms to the left:

0-(24x^2+71x-16)=0

We add all the numbers together, and all the variables

-(24x^2+71x-16)=0

We get rid of parentheses

-24x^2-71x+16=0

a = -24; b = -71; c = +16;
Δ = b2-4ac
Δ = -712-4·(-24)·16
Δ = 6577
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-71)-\sqrt{6577}}{2*-24}=\frac{71-\sqrt{6577}}{-48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-71)+\sqrt{6577}}{2*-24}=\frac{71+\sqrt{6577}}{-48}
$