Use the quadratic formula to solve the equation. -2x^2+3x+5=0

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Use the quadratic formula to solve the equation. -2x^2+3x+5=0

Algebra
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Compare your quadratic equation with \(ax^2+bx+c=0\) find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\] then the two roots of x are: \(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
somethings not coming up correctly :(. I tried plugging everything in.
can you plz show your work?

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\[\sqrt{3x^2-4(2x^2*5)}\]
okk...we can't have x in the values of a,b,c |dw:1442238245129:dw|
a = -2, b =3 c = 5 got it? can you now try again ?
i tried again still don't have anything :/ i'm trying to put it in the TI-30XS calculator.
start by finding b^2 -4ac (3)^2 - 4(-2)(5) = ... ?
49
and its square root is ?
7 because 7*7 is 49 :D
yes! now we have the \(\sqrt{b^2 -4ac}\) part which equals 7
\(numerator = -b \pm \sqrt{b^2-4ac} = -3 \pm 7\) so what the the 2 values of numerator? -3 +7 = ...? -3-7 = ... ?
4 & -10
yes, now the denominator 2a = 2(-2) = -4 so your 2 values of x are -4/4 and -10/4 simplify them!
Let's solve your equation step-by-step. −2x2+3x+5=0 Step 1: Factor left side of equation. (−2x+5)(x+1)=0 Step 2: Set factors equal to 0. −2x+5=0 or x+1=0 x= 5 2 or x=−1 Answer: x= 5 2 or x=−1
-1 & -5/2?
very good! they're correct :)
Yes! thank you so much for all your help :).
actually, my bad
the 2nd numerator is -10 so the 2nd value is -10/-4 = 5/2
-1 and 5/2
but you understand the steps, right? :)

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