4b^2+8b-35=0

Solution for 4b^2+8b-35=0 equation:

4b^2+8b-35=0

a = 4; b = 8; c = -35;
Δ = b2-4ac
Δ = 82-4·4·(-35)
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{39}}{2*4}=\frac{-8-4\sqrt{39}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{39}}{2*4}=\frac{-8+4\sqrt{39}}{8}
$