21m^2+98m-59=5m+1

Solution for 21m^2+98m-59=5m+1 equation:

21m^2+98m-59=5m+1

We move all terms to the left:

21m^2+98m-59-(5m+1)=0

We get rid of parentheses

21m^2+98m-5m-1-59=0

We add all the numbers together, and all the variables

21m^2+93m-60=0

a = 21; b = 93; c = -60;
Δ = b2-4ac
Δ = 932-4·21·(-60)
Δ = 13689
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{13689}=117$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(93)-117}{2*21}=\frac{-210}{42}
=-5
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(93)+117}{2*21}=\frac{24}{42}
=4/7
$