R(10.5)=-10p^2+130p

Solution for R(10.5)=-10p^2+130p equation:

(10.5)=-10R^2+130R

We move all terms to the left:

(10.5)-(-10R^2+130R)=0

We add all the numbers together, and all the variables

-(-10R^2+130R)+10.5=0

We get rid of parentheses

10R^2-130R+10.5=0

a = 10; b = -130; c = +10.5;
Δ = b2-4ac
Δ = -1302-4·10·10.5
Δ = 16480
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16480}=\sqrt{16*1030}=\sqrt{16}*\sqrt{1030}=4\sqrt{1030}$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-130)-4\sqrt{1030}}{2*10}=\frac{130-4\sqrt{1030}}{20}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-130)+4\sqrt{1030}}{2*10}=\frac{130+4\sqrt{1030}}{20}
$