6x^2+41x=7

Solution for 6x^2+41x=7 equation:

6x^2+41x=7

We move all terms to the left:

6x^2+41x-(7)=0

a = 6; b = 41; c = -7;
Δ = b2-4ac
Δ = 412-4·6·(-7)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-43}{2*6}=\frac{-84}{12}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+43}{2*6}=\frac{2}{12}
=1/6
$