10(3x^2-13x)+40=0

Solution for 10(3x^2-13x)+40=0 equation:

10(3x^2-13x)+40=0

We multiply parentheses

30x^2-130x+40=0

a = 30; b = -130; c = +40;
Δ = b2-4ac
Δ = -1302-4·30·40
Δ = 12100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12100}=110$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-130)-110}{2*30}=\frac{20}{60}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-130)+110}{2*30}=\frac{240}{60}
=4
$