4x^2+18=98

Solution for 4x^2+18=98 equation:

4x^2+18=98

We move all terms to the left:

4x^2+18-(98)=0

We add all the numbers together, and all the variables

4x^2-80=0

a = 4; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·4·(-80)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*4}=\frac{0-16\sqrt{5}}{8}
=-\frac{16\sqrt{5}}{8}
=-2\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*4}=\frac{0+16\sqrt{5}}{8}
=\frac{16\sqrt{5}}{8}
=2\sqrt{5}
$