4r+r^2=98

Solution for 4r+r^2=98 equation:

4r+r^2=98

We move all terms to the left:

4r+r^2-(98)=0

a = 1; b = 4; c = -98;
Δ = b2-4ac
Δ = 42-4·1·(-98)
Δ = 408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{102}}{2*1}=\frac{-4-2\sqrt{102}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{102}}{2*1}=\frac{-4+2\sqrt{102}}{2}
$