-7b^2+18=0

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

-7b^2+18=0

a = -7; b = 0; c = +18;
Δ = b2-4ac
Δ = 02-4·(-7)·18
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*-7}=\frac{0-6\sqrt{14}}{-14}
=-\frac{6\sqrt{14}}{-14}
=-\frac{3\sqrt{14}}{-7}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*-7}=\frac{0+6\sqrt{14}}{-14}
=\frac{6\sqrt{14}}{-14}
=\frac{3\sqrt{14}}{-7}
$