P(b)=5/2b+45

Solution for P(b)=5/2b+45 equation:

(P)=5/2P+45

We move all terms to the left:

(P)-(5/2P+45)=0


Domain of the equation: 2P+45)!=0

P∈R


We get rid of parentheses

P-5/2P-45=0

We multiply all the terms by the denominator


P*2P-45*2P-5=0

Wy multiply elements

2P^2-90P-5=0

a = 2; b = -90; c = -5;
Δ = b2-4ac
Δ = -902-4·2·(-5)
Δ = 8140
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8140}=\sqrt{4*2035}=\sqrt{4}*\sqrt{2035}=2\sqrt{2035}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-2\sqrt{2035}}{2*2}=\frac{90-2\sqrt{2035}}{4}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+2\sqrt{2035}}{2*2}=\frac{90+2\sqrt{2035}}{4}
$