10x^2=41x+45

Solution for 10x^2=41x+45 equation:

10x^2=41x+45

We move all terms to the left:

10x^2-(41x+45)=0

We get rid of parentheses

10x^2-41x-45=0

a = 10; b = -41; c = -45;
Δ = b2-4ac
Δ = -412-4·10·(-45)
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-59}{2*10}=\frac{-18}{20}
=-9/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+59}{2*10}=\frac{100}{20}
=5
$