6(8+3c)+4(10+2c)c=2

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Solution for 6(8+3c)+4(10+2c)c=2 equation:



6(8+3c)+4(10+2c)c=2
We move all terms to the left:
6(8+3c)+4(10+2c)c-(2)=0
We add all the numbers together, and all the variables
6(3c+8)+4(2c+10)c-2=0
We multiply parentheses
8c^2+18c+40c+48-2=0
We add all the numbers together, and all the variables
8c^2+58c+46=0
a = 8; b = 58; c = +46;
Δ = b2-4ac
Δ = 582-4·8·46
Δ = 1892
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1892}=\sqrt{4*473}=\sqrt{4}*\sqrt{473}=2\sqrt{473}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-2\sqrt{473}}{2*8}=\frac{-58-2\sqrt{473}}{16} $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+2\sqrt{473}}{2*8}=\frac{-58+2\sqrt{473}}{16} $

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