(5x+10)(3x+4)=30

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Solution for (5x+10)(3x+4)=30 equation:



(5x+10)(3x+4)=30
We move all terms to the left:
(5x+10)(3x+4)-(30)=0
We multiply parentheses ..
(+15x^2+20x+30x+40)-30=0
We get rid of parentheses
15x^2+20x+30x+40-30=0
We add all the numbers together, and all the variables
15x^2+50x+10=0
a = 15; b = 50; c = +10;
Δ = b2-4ac
Δ = 502-4·15·10
Δ = 1900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{19}}{2*15}=\frac{-50-10\sqrt{19}}{30} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{19}}{2*15}=\frac{-50+10\sqrt{19}}{30} $

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