10x^2+19x-6=9

Solution for 10x^2+19x-6=9 equation:

10x^2+19x-6=9

We move all terms to the left:

10x^2+19x-6-(9)=0

We add all the numbers together, and all the variables

10x^2+19x-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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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