The solution set for the both equation is (3,0) and (-1,0)
We have given that the equation
[tex]y=x^2-2x-3 \ and\ y=x+3[/tex]
We have to determine the solution to the first equation
[tex]\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-3\right)}}{2\cdot \:1}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-2\right)\pm \:4}{2\cdot \:1}[/tex]
[tex]x_1=\frac{-\left(-2\right)+4}{2\cdot \:1},\:x_2=\frac{-\left(-2\right)-4}{2\cdot \:1}[/tex]
[tex]x=3,\:x=-1[/tex]
Now use this value we have to find the value of y
When x=3 then
y=(3)^2-2(3)-3
y=9-6-3
y=0
When x=-1 then y=1-2(-1)-3
y=1+2-3
y=0
Therefore the solution set for both equations is (3,0) and (-1,0).
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