7r^2+5r-18=0

Solution for 7r^2+5r-18=0 equation:

7r^2+5r-18=0

a = 7; b = 5; c = -18;
Δ = b2-4ac
Δ = 52-4·7·(-18)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-23}{2*7}=\frac{-28}{14}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+23}{2*7}=\frac{18}{14}
=1+2/7
$