28x^2+21x-10=20x+5

Solution for 28x^2+21x-10=20x+5 equation:

28x^2+21x-10=20x+5

We move all terms to the left:

28x^2+21x-10-(20x+5)=0

We get rid of parentheses

28x^2+21x-20x-5-10=0

We add all the numbers together, and all the variables

28x^2+x-15=0

a = 28; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·28·(-15)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*28}=\frac{-42}{56}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*28}=\frac{40}{56}
=5/7
$