7x^2-38=5x

Solution for 7x^2-38=5x equation:

7x^2-38=5x

We move all terms to the left:

7x^2-38-(5x)=0

a = 7; b = -5; c = -38;
Δ = b2-4ac
Δ = -52-4·7·(-38)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-33}{2*7}=\frac{-28}{14}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+33}{2*7}=\frac{38}{14}
=2+5/7
$