(2b2+2)/10+4=17

Solution for (2b2+2)/10+4=17 equation:

(2b^2+2)/10+4=17

We move all terms to the left:

(2b^2+2)/10+4-(17)=0

We add all the numbers together, and all the variables

(2b^2+2)/10-13=0

We multiply all the terms by the denominator


(2b^2+2)-13*10=0

We add all the numbers together, and all the variables

(2b^2+2)-130=0

We get rid of parentheses

2b^2+2-130=0

We add all the numbers together, and all the variables

2b^2-128=0

a = 2; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·2·(-128)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*2}=\frac{-32}{4}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*2}=\frac{32}{4}
=8
$