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Mrbonez321
Solve for x: 2x^2 - 4x - 14 = 0 I have everything up to 2(x-2)^2=6, but I do not know how to continue.
2(x^2 - 2x - 7) = 0 Δ = 4 - ( -28) = 32 x = (-4 ± 4 √2)/ 4 = -1 ± √2
Oops, x = (-4 ± 4 √2)/ 2 = -2 ± 2√2
I dont understand Δ... I am only in Algebra II
2x^2 - 4x - 14 = 0 divide everything by 2 in order to get x^2 by itself without the coefficient x^2 - 2x - 7 = 0 and then use the quadratic formula (-b+-sqrt(b^2 - 4ac))/2a (2 +or- sqrt(4 - 4*1*-7))/2 (2 + sqrt(32)) / 2 and (2 - sqrt(32))/2 2 + 4sqrt(2) / 2 and 2 - 4sqrt(2) / 2 1 + 2sqrt(2) and 1 - 2sqrt(2)
That's quadratic formula in algebra II
Only quadratic formula can solve any cases!
S's result is perfect, not typo as mine!