4b^2+10b-6=0

Solution for 4b^2+10b-6=0 equation:

4b^2+10b-6=0

a = 4; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·4·(-6)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*4}=\frac{-24}{8}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*4}=\frac{4}{8}
=1/2
$