R=1.03+7.11t-0.38t^2

Solution for R=1.03+7.11t-0.38t^2 equation:

=1.03+7.11R-0.38R^2

We move all terms to the left:

-(1.03+7.11R-0.38R^2)=0

We get rid of parentheses

0.38R^2-7.11R-1.03=0

a = 0.38; b = -7.11; c = -1.03;
Δ = b2-4ac
Δ = -7.112-4·0.38·(-1.03)
Δ = 52.1177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7.11)-\sqrt{52.1177}}{2*0.38}=\frac{7.11-\sqrt{52.1177}}{0.76}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7.11)+\sqrt{52.1177}}{2*0.38}=\frac{7.11+\sqrt{52.1177}}{0.76}
$