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.5x^2+2x-266=0
a = .5; b = 2; c = -266;
Δ = b2-4ac
Δ = 22-4·.5·(-266)
Δ = 536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{134}}{2*.5}=\frac{-2-2\sqrt{134}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{134}}{2*.5}=\frac{-2+2\sqrt{134}}{1} $
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