r2+3r-8=0

Solution for r2+3r-8=0 equation:

r2+3r-8=0

We add all the numbers together, and all the variables

r^2+3r-8=0

a = 1; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·1·(-8)
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*1}=\frac{-3-\sqrt{41}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*1}=\frac{-3+\sqrt{41}}{2}
$