[tex]x=2+½*\sqrt{10} [/tex]
or
[tex]x=2-½*\sqrt{10} [/tex]
Answer:
Solution given:
y=2x²-5x+6.....[1]
y=3x+3.........….[2]
solving equation 1&2
2x²-5x+6=3x+3
2x²-5x-3x+6-3=0
2x²-8x+3=0
comparing above equation with ax²+bx+c=0 we get
a=2
b=-8
c=3
by using quadratic equation
[tex]x=\frac{+8±\sqrt{(-8)²-4*2*3}}{2*2}[/tex]
[tex]x=\frac{+8±2\sqrt{10}}{4}[/tex]
taking positive
[tex]x=\frac{+8+2\sqrt{10}}{4}[/tex]
[tex]x=2+½*\sqrt{10} [/tex]
taking negative
[tex]x=2-½*\sqrt{10} [/tex]
