3x^2+17x=20

Solution for 3x^2+17x=20 equation:

3x^2+17x=20

We move all terms to the left:

3x^2+17x-(20)=0

a = 3; b = 17; c = -20;
Δ = b2-4ac
Δ = 172-4·3·(-20)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*3}=\frac{-40}{6}
=-6+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*3}=\frac{6}{6}
=1
$