2x^2+3x+10=2(2x+16)

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

2x^2+3x+10=2(2x+16)

We move all terms to the left:

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


We calculate terms in parentheses: -(2(2x+16)), so:

2(2x+16)

We multiply parentheses

4x+32

Back to the equation:

-(4x+32)


We get rid of parentheses

2x^2+3x-4x-32+10=0

We add all the numbers together, and all the variables

2x^2-1x-22=0

a = 2; b = -1; c = -22;
Δ = b2-4ac
Δ = -12-4·2·(-22)
Δ = 177
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{-(-1)-\sqrt{177}}{2*2}=\frac{1-\sqrt{177}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{177}}{2*2}=\frac{1+\sqrt{177}}{4}
$