Q=10x^2+19x-15=0

Solution for Q=10x^2+19x-15=0 equation:

=10Q^2+19Q-15=0

We move all terms to the left:

-(10Q^2+19Q-15)=0

We get rid of parentheses

-10Q^2-19Q+15=0

a = -10; b = -19; c = +15;
Δ = b2-4ac
Δ = -192-4·(-10)·15
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-31}{2*-10}=\frac{-12}{-20}
=3/5
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+31}{2*-10}=\frac{50}{-20}
=-2+1/2
$