Respuesta :
Answer:
The solution of the quadratic equation is, (-6.583, 2.583).
Step-by-step explanation:
The equation provided is: [tex](3-y)(y+4)=3y-5[/tex]
Simplify the equation as follows:
[tex](3-y)(y+4) = 3y-5\\3y + 12 - y^{2} - 4y = 3y - 5\\0 = y^{2} + 4y - 17[/tex]
The resultant quadratic equation is not a perfect square.
So use the quadratic formula to compute the roots:
[tex]y=[\frac{-b-\sqrt{b^{2} -4ac} }{2a}, \frac{-b+\sqrt{b^{2} -4ac} }{2a}][/tex]
Here, a = 1, b = 4 and c = -17
The roots are:
[tex]y=[\frac{-b-\sqrt{b^{2}-4ac} }{2a}, \frac{-b+\sqrt{b^{2} -4ac} }{2a}]\\=[\frac{-4-\sqrt{4^{2}-( 4\times1\times(-17))} }{2\times1}, \frac{-4+\sqrt{4^{2}-( 4\times1\times(-17))} }{2\times1}]\\=[\frac{-4-2\sqrt{21} }{2}, \frac{-4+2\sqrt{21} }{2}]\\=[-2-\sqrt{21}, -2+\sqrt{21]}[/tex]
The value of [tex]\sqrt{21}[/tex] is approximately 4.583.
So the roots of the equation are:
[tex]y=[-2-\sqrt{21}, -2+\sqrt{21]}\\=[-2-4.583, -2+4.583]\\=[-6.583, 2.583][/tex]
