4x^2-10x=19

Solution for 4x^2-10x=19 equation:

4x^2-10x=19

We move all terms to the left:

4x^2-10x-(19)=0

a = 4; b = -10; c = -19;
Δ = b2-4ac
Δ = -102-4·4·(-19)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{101}}{2*4}=\frac{10-2\sqrt{101}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{101}}{2*4}=\frac{10+2\sqrt{101}}{8}
$