11(r^2+4)+28r=5r(2r+3)+14

Solution for 11(r^2+4)+28r=5r(2r+3)+14 equation:

11(r^2+4)+28r=5r(2r+3)+14

We move all terms to the left:

11(r^2+4)+28r-(5r(2r+3)+14)=0

We add all the numbers together, and all the variables

28r+11(r^2+4)-(5r(2r+3)+14)=0

We multiply parentheses

11r^2+28r-(5r(2r+3)+14)+44=0


We calculate terms in parentheses: -(5r(2r+3)+14), so:

5r(2r+3)+14

We multiply parentheses

10r^2+15r+14

Back to the equation:

-(10r^2+15r+14)


We get rid of parentheses

11r^2-10r^2+28r-15r-14+44=0

We add all the numbers together, and all the variables

r^2+13r+30=0

a = 1; b = 13; c = +30;
Δ = b2-4ac
Δ = 132-4·1·30
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*1}=\frac{-20}{2}
=-10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*1}=\frac{-6}{2}
=-3
$