0.75x^2+19x-25=0

Solution for 0.75x^2+19x-25=0 equation:

0.75x^2+19x-25=0

a = 0.75; b = 19; c = -25;
Δ = b2-4ac
Δ = 192-4·0.75·(-25)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-2\sqrt{109}}{2*0.75}=\frac{-19-2\sqrt{109}}{1.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+2\sqrt{109}}{2*0.75}=\frac{-19+2\sqrt{109}}{1.5}
$