2x^2+3x+10=4x+16

Solution for 2x^2+3x+10=4x+16 equation:

2x^2+3x+10=4x+16

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-1x-6=0

a = 2; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·2·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*2}=\frac{-6}{4}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*2}=\frac{8}{4}
=2
$