2r2+-3r-4=0

Solution for 2r2+-3r-4=0 equation:

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

We add all the numbers together, and all the variables

2r^2-3r=0

a = 2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*2}=\frac{0}{4}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*2}=\frac{6}{4}
=1+1/2
$