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solve \[2x ^{2}-8x=-7\]
add 7 to both sides. \(2x^2 - 8x +7=0\) then apply the general formula.
\[\frac{ -(-8)\sqrt{8x ^{2}-2x*7} }{ 2*2 }\]
it's \(x=\frac{-(-8)\pm \sqrt{(-8)^2-4(2)(7)}}{2(2)}\) though
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ok so simplify all that its \[\frac{ 8\sqrt{64-14} }{ 4 }\]
u forgot to multiply the 14 by 4? the others are right
so \[\frac{ 8\sqrt{64-56} }{ 4 }\]
yup! you've got it. :)
Its soo easy :D
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loool
so simplify that its \[\frac{ 2\sqrt{2} }{ 2 }\]
wait. you've missed the plus minus out. \(\frac{8 \pm \sqrt{8}}{4}\)=\(\frac{8 \pm 2\sqrt{2}}{4}=\frac{4 \pm \sqrt{2}}{2}\)
ok thnx
you’re welcome :)
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