x2+21x=98

Solution for x2+21x=98 equation:

x2+21x=98

We move all terms to the left:

x2+21x-(98)=0

We add all the numbers together, and all the variables

x^2+21x-98=0

a = 1; b = 21; c = -98;
Δ = b2-4ac
Δ = 212-4·1·(-98)
Δ = 833
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{833}=\sqrt{49*17}=\sqrt{49}*\sqrt{17}=7\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-7\sqrt{17}}{2*1}=\frac{-21-7\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+7\sqrt{17}}{2*1}=\frac{-21+7\sqrt{17}}{2}
$