((5x)^2)-19x+12=0

Solution for ((5x)^2)-19x+12=0 equation:

((5x)^2)-19x+12=0

determiningTheFunctionDomain
5x^2-19x+12=0

a = 5; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·5·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*5}=\frac{8}{10}
=4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*5}=\frac{30}{10}
=3
$