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0=2x^2+10x+5
We move all terms to the left:
0-(2x^2+10x+5)=0
We add all the numbers together, and all the variables
-(2x^2+10x+5)=0
We get rid of parentheses
-2x^2-10x-5=0
a = -2; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·(-2)·(-5)
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{15}}{2*-2}=\frac{10-2\sqrt{15}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{15}}{2*-2}=\frac{10+2\sqrt{15}}{-4} $
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