4x^2+10x-28=0

Solution for 4x^2+10x-28=0 equation:

4x^2+10x-28=0

a = 4; b = 10; c = -28;
Δ = b2-4ac
Δ = 102-4·4·(-28)
Δ = 548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{137}}{2*4}=\frac{-10-2\sqrt{137}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{137}}{2*4}=\frac{-10+2\sqrt{137}}{8}
$