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4.9x^2-20x-10=0
a = 4.9; b = -20; c = -10;
Δ = b2-4ac
Δ = -202-4·4.9·(-10)
Δ = 596
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{596}=\sqrt{4*149}=\sqrt{4}*\sqrt{149}=2\sqrt{149}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{149}}{2*4.9}=\frac{20-2\sqrt{149}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{149}}{2*4.9}=\frac{20+2\sqrt{149}}{9.8} $
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