1b2-3=10b+9

Solution for 1b2-3=10b+9 equation:

1b^2-3=10b+9

We move all terms to the left:

1b^2-3-(10b+9)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

b^2-10b-9-3=0

We add all the numbers together, and all the variables

b^2-10b-12=0

a = 1; b = -10; c = -12;
Δ = b2-4ac
Δ = -102-4·1·(-12)
Δ = 148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{37}}{2*1}=\frac{10-2\sqrt{37}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{37}}{2*1}=\frac{10+2\sqrt{37}}{2}
$