b^2+35=175

Solution for b^2+35=175 equation:

b^2+35=175

We move all terms to the left:

b^2+35-(175)=0

We add all the numbers together, and all the variables

b^2-140=0

a = 1; b = 0; c = -140;
Δ = b2-4ac
Δ = 02-4·1·(-140)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*1}=\frac{0-4\sqrt{35}}{2}
=-\frac{4\sqrt{35}}{2}
=-2\sqrt{35}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*1}=\frac{0+4\sqrt{35}}{2}
=\frac{4\sqrt{35}}{2}
=2\sqrt{35}
$