x^2-21x+98=)

Solution for x^2-21x+98=) equation:

x^2-21x+98=)

We move all terms to the left:

x^2-21x+98-())=0

We add all the numbers together, and all the variables

x^2-21x=0

a = 1; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·1·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*1}=\frac{42}{2}
=21
$