42x^2=29x+5

Solution for 42x^2=29x+5 equation:

42x^2=29x+5

We move all terms to the left:

42x^2-(29x+5)=0

We get rid of parentheses

42x^2-29x-5=0

a = 42; b = -29; c = -5;
Δ = b2-4ac
Δ = -292-4·42·(-5)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-41}{2*42}=\frac{-12}{84}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+41}{2*42}=\frac{70}{84}
=5/6
$