4a^2+14a-3=0

Solution for 4a^2+14a-3=0 equation:

4a^2+14a-3=0

a = 4; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·4·(-3)
Δ = 244
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{61}}{2*4}=\frac{-14-2\sqrt{61}}{8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{61}}{2*4}=\frac{-14+2\sqrt{61}}{8}
$