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25x^2+40x+16=48
We move all terms to the left:
25x^2+40x+16-(48)=0
We add all the numbers together, and all the variables
25x^2+40x-32=0
a = 25; b = 40; c = -32;
Δ = b2-4ac
Δ = 402-4·25·(-32)
Δ = 4800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{3}}{2*25}=\frac{-40-40\sqrt{3}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{3}}{2*25}=\frac{-40+40\sqrt{3}}{50} $
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