0=2r^2+3r-4

Solution for 0=2r^2+3r-4 equation:

0=2r^2+3r-4

We move all terms to the left:

0-(2r^2+3r-4)=0

We add all the numbers together, and all the variables

-(2r^2+3r-4)=0

We get rid of parentheses

-2r^2-3r+4=0

a = -2; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-2)·4
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*-2}=\frac{3-\sqrt{41}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*-2}=\frac{3+\sqrt{41}}{-4}
$