3x2+20x=7/3

Solution for 3x2+20x=7/3 equation:

3x^2+20x=7/3

We move all terms to the left:

3x^2+20x-(7/3)=0

We add all the numbers together, and all the variables

3x^2+20x-(+7/3)=0

We get rid of parentheses

3x^2+20x-7/3=0

We multiply all the terms by the denominator


3x^2*3+20x*3-7=0

Wy multiply elements

9x^2+60x-7=0

a = 9; b = 60; c = -7;
Δ = b2-4ac
Δ = 602-4·9·(-7)
Δ = 3852
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:

$\sqrt{\Delta}=\sqrt{3852}=\sqrt{36*107}=\sqrt{36}*\sqrt{107}=6\sqrt{107}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-6\sqrt{107}}{2*9}=\frac{-60-6\sqrt{107}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+6\sqrt{107}}{2*9}=\frac{-60+6\sqrt{107}}{18}
$