1.5r2+4.5+6r=-0.3-3r+r2

Solution for 1.5r2+4.5+6r=-0.3-3r+r2 equation:

1.5r^2+4.5+6r=-0.3-3r+r2

We move all terms to the left:

1.5r^2+4.5+6r-(-0.3-3r+r2)=0

We add all the numbers together, and all the variables

1.5r^2-(-3r+r^2-0.3)+6r+4.5=0

We get rid of parentheses

1.5r^2-r^2+3r+6r+0.3+4.5=0

We add all the numbers together, and all the variables

0.5r^2+9r+4.8=0

a = 0.5; b = 9; c = +4.8;
Δ = b2-4ac
Δ = 92-4·0.5·4.8
Δ = 71.4
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{-(9)-\sqrt{71.4}}{2*0.5}=\frac{-9-\sqrt{71.4}}{1}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{71.4}}{2*0.5}=\frac{-9+\sqrt{71.4}}{1}
$