5b^2+9b-18=3+10b

Solution for 5b^2+9b-18=3+10b equation:

5b^2+9b-18=3+10b

We move all terms to the left:

5b^2+9b-18-(3+10b)=0

We add all the numbers together, and all the variables

5b^2+9b-(10b+3)-18=0

We get rid of parentheses

5b^2+9b-10b-3-18=0

We add all the numbers together, and all the variables

5b^2-1b-21=0

a = 5; b = -1; c = -21;
Δ = b2-4ac
Δ = -12-4·5·(-21)
Δ = 421
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{421}}{2*5}=\frac{1-\sqrt{421}}{10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{421}}{2*5}=\frac{1+\sqrt{421}}{10}
$