12x+16=18x^2

Solution for 12x+16=18x^2 equation:

12x+16=18x^2

We move all terms to the left:

12x+16-(18x^2)=0

determiningTheFunctionDomain
-18x^2+12x+16=0

a = -18; b = 12; c = +16;
Δ = b2-4ac
Δ = 122-4·(-18)·16
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*-18}=\frac{-48}{-36}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*-18}=\frac{24}{-36}
=-2/3
$