20x^2=14+27x

Solution for 20x^2=14+27x equation:

20x^2=14+27x

We move all terms to the left:

20x^2-(14+27x)=0

We add all the numbers together, and all the variables

20x^2-(27x+14)=0

We get rid of parentheses

20x^2-27x-14=0

a = 20; b = -27; c = -14;
Δ = b2-4ac
Δ = -272-4·20·(-14)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-43}{2*20}=\frac{-16}{40}
=-2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+43}{2*20}=\frac{70}{40}
=1+3/4
$