Seudónimo
Seudónimo
02-12-2016
Mathematics
contestada
use the quadratic formula to solve 5x^2+2x-3
Respuesta :
apologiabiology
apologiabiology
02-12-2016
assuming it equals 0
for
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
given
5x^2+2x-3=0
a=5
b=2
c=-3
x=[tex] \frac{-2+/- \sqrt{2^2-4(5)(-3)} }{2(5)} [/tex]
x=[tex] \frac{-2+/- \sqrt{4+60} }{10} [/tex]
x=[tex] \frac{-2+/- \sqrt{64} }{10} [/tex]
x=[tex] \frac{-2+/- 8 }{10} [/tex]
x=(-2+8)/10 or (-2-8)/10
x=6/10 or -10/10
x=3/5 or -1
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