4x^2+10x=38

Solution for 4x^2+10x=38 equation:

4x^2+10x=38

We move all terms to the left:

4x^2+10x-(38)=0

a = 4; b = 10; c = -38;
Δ = b2-4ac
Δ = 102-4·4·(-38)
Δ = 708
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:

$\sqrt{\Delta}=\sqrt{708}=\sqrt{4*177}=\sqrt{4}*\sqrt{177}=2\sqrt{177}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{177}}{2*4}=\frac{-10-2\sqrt{177}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{177}}{2*4}=\frac{-10+2\sqrt{177}}{8}
$