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

Solution for 2b^2+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

determiningTheFunctionDomain
2b^2+4-17+2/10=0

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

2b^2*10-128=0

Wy multiply elements

20b^2-128=0

a = 20; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·20·(-128)
Δ = 10240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{10240}=\sqrt{1024*10}=\sqrt{1024}*\sqrt{10}=32\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{10}}{2*20}=\frac{0-32\sqrt{10}}{40}
=-\frac{32\sqrt{10}}{40}
=-\frac{4\sqrt{10}}{5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{10}}{2*20}=\frac{0+32\sqrt{10}}{40}
=\frac{32\sqrt{10}}{40}
=\frac{4\sqrt{10}}{5}
$