0=4(3x^2-110x+600)

Solution for 0=4(3x^2-110x+600) equation:

0=4(3x^2-110x+600)

We move all terms to the left:

0-(4(3x^2-110x+600))=0

We add all the numbers together, and all the variables

-(4(3x^2-110x+600))=0


We calculate terms in parentheses: -(4(3x^2-110x+600)), so:

4(3x^2-110x+600)

We multiply parentheses

12x^2-440x+2400

Back to the equation:

-(12x^2-440x+2400)


We get rid of parentheses

-12x^2+440x-2400=0

a = -12; b = 440; c = -2400;
Δ = b2-4ac
Δ = 4402-4·(-12)·(-2400)
Δ = 78400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{78400}=280$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(440)-280}{2*-12}=\frac{-720}{-24}
=+30
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(440)+280}{2*-12}=\frac{-160}{-24}
=6+2/3
$