x^2+10+3x^2+10x+15=3x^2+16

Solution for x^2+10+3x^2+10x+15=3x^2+16 equation:

x^2+10+3x^2+10x+15=3x^2+16

We move all terms to the left:

x^2+10+3x^2+10x+15-(3x^2+16)=0

We add all the numbers together, and all the variables

4x^2+10x-(3x^2+16)+25=0

We get rid of parentheses

4x^2-3x^2+10x-16+25=0

We add all the numbers together, and all the variables

x^2+10x+9=0

a = 1; b = 10; c = +9;
Δ = b2-4ac
Δ = 102-4·1·9
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*1}=\frac{-2}{2}
=-1
$