(16x^2+22x-3)=0

Solution for (16x^2+22x-3)=0 equation:

(16x^2+22x-3)=0

We get rid of parentheses

16x^2+22x-3=0

a = 16; b = 22; c = -3;
Δ = b2-4ac
Δ = 222-4·16·(-3)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*16}=\frac{-48}{32}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*16}=\frac{4}{32}
=1/8
$