The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)
Solution:
Given that,
[tex]y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2[/tex]
We have to substitute eqn 1 in eqn 2
[tex]x + 3 = x^2 + 2x - 4[/tex]
[tex]\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0[/tex]
[tex]\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}[/tex]
[tex]x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}[/tex]
[tex]x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925[/tex]
[tex]Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925[/tex]
Substitute x = 2.1925 in eqn 1
y = 2.1925 + 3
y = 5.1925
Substitute x = -3.1925 in eqn 1
y = -3.1925 + 3
y = -0.1925
Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)
