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2y^2-8y+5=0 <--- solve by the quadratic formula , set the equation to "0" first. must be in simplest radical form, show work please
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If I solve it, it will only confuse you. I have this crazy idea that all quadratics are factorable
ohh:p
-8+\[8\pm \sqrt{(-8)^2-4(2)(5)}\] Im doing the numerator first
You could have at least posted the actual quadratic formula first
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you solve the inside of square root you will get \[8\pm \sqrt{64-40}\]
\[8\pm \sqrt{24}\]
\[8\pm2\sqrt{6}\]
now all of this is over the denominator 2(2) correct it simplifies to \[2\pm (\sqrt{6}\div2)\]
thanks
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