105=(3x^2+21x+30)*0.5

Solution for 105=(3x^2+21x+30)*0.5 equation:

105=(3x^2+21x+30)*0.5

We move all terms to the left:

105-((3x^2+21x+30)*0.5)=0


We calculate terms in parentheses: -((3x^2+21x+30)*0.5), so:

(3x^2+21x+30)*0.5

We multiply parentheses

1.5x^2+10.5x+15

Back to the equation:

-(1.5x^2+10.5x+15)


We get rid of parentheses

-1.5x^2-10.5x-15+105=0

We add all the numbers together, and all the variables

-1.5x^2-10.5x+90=0

a = -1.5; b = -10.5; c = +90;
Δ = b2-4ac
Δ = -10.52-4·(-1.5)·90
Δ = 650.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10.5)-\sqrt{650.25}}{2*-1.5}=\frac{10.5-\sqrt{650.25}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10.5)+\sqrt{650.25}}{2*-1.5}=\frac{10.5+\sqrt{650.25}}{-3}
$