10m^2+7m=3

Solution for 10m^2+7m=3 equation:

10m^2+7m=3

We move all terms to the left:

10m^2+7m-(3)=0

a = 10; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·10·(-3)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*10}=\frac{-20}{20}
=-1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*10}=\frac{6}{20}
=3/10
$