3x^2-18x+(96/13)=0

Solution for 3x^2-18x+(96/13)=0 equation:

3x^2-18x+(96/13)=0

We add all the numbers together, and all the variables

3x^2-18x+(+96/13)=0

We get rid of parentheses

3x^2-18x+96/13=0

We multiply all the terms by the denominator


3x^2*13-18x*13+96=0

Wy multiply elements

39x^2-234x+96=0

a = 39; b = -234; c = +96;
Δ = b2-4ac
Δ = -2342-4·39·96
Δ = 39780
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{39780}=\sqrt{36*1105}=\sqrt{36}*\sqrt{1105}=6\sqrt{1105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-234)-6\sqrt{1105}}{2*39}=\frac{234-6\sqrt{1105}}{78}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-234)+6\sqrt{1105}}{2*39}=\frac{234+6\sqrt{1105}}{78}
$