-28x^2+8x+3=0

Solution for -28x^2+8x+3=0 equation:

-28x^2+8x+3=0

a = -28; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·(-28)·3
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*-28}=\frac{-28}{-56}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*-28}=\frac{12}{-56}
=-3/14
$