7x^2=6+41x

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

7x^2=6+41x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

7x^2-41x-6=0

a = 7; b = -41; c = -6;
Δ = b2-4ac
Δ = -412-4·7·(-6)
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*7}=\frac{-2}{14}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+43}{2*7}=\frac{84}{14}
=6
$