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2x(4x-10)=98
We move all terms to the left:
2x(4x-10)-(98)=0
We multiply parentheses
8x^2-20x-98=0
a = 8; b = -20; c = -98;
Δ = b2-4ac
Δ = -202-4·8·(-98)
Δ = 3536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3536}=\sqrt{16*221}=\sqrt{16}*\sqrt{221}=4\sqrt{221}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{221}}{2*8}=\frac{20-4\sqrt{221}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{221}}{2*8}=\frac{20+4\sqrt{221}}{16} $
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