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5x^2+10x-16=-12
We move all terms to the left:
5x^2+10x-16-(-12)=0
We add all the numbers together, and all the variables
5x^2+10x-4=0
a = 5; b = 10; c = -4;
Δ = b2-4ac
Δ = 102-4·5·(-4)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{5}}{2*5}=\frac{-10-6\sqrt{5}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{5}}{2*5}=\frac{-10+6\sqrt{5}}{10} $
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