9x^2-42x+49=9/16

Solution for 9x^2-42x+49=9/16 equation:

9x^2-42x+49=9/16

We move all terms to the left:

9x^2-42x+49-(9/16)=0

We add all the numbers together, and all the variables

9x^2-42x+49-(+9/16)=0

We get rid of parentheses

9x^2-42x+49-9/16=0

We multiply all the terms by the denominator


9x^2*16-42x*16-9+49*16=0

We add all the numbers together, and all the variables

9x^2*16-42x*16+775=0

Wy multiply elements

144x^2-672x+775=0

a = 144; b = -672; c = +775;
Δ = b2-4ac
Δ = -6722-4·144·775
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-672)-72}{2*144}=\frac{600}{288}
=2+1/12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-672)+72}{2*144}=\frac{744}{288}
=2+7/12
$