4w^2+10w-3=0

Solution for 4w^2+10w-3=0 equation:

4w^2+10w-3=0

a = 4; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·4·(-3)
Δ = 148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{37}}{2*4}=\frac{-10-2\sqrt{37}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{37}}{2*4}=\frac{-10+2\sqrt{37}}{8}
$