Answer:
x = 5, -7
Step-by-step explanation:
[tex]x^2+2x-35=0[/tex]
[tex]\mathrm{Solve\;with\;quadratic\;formula}[/tex]
[tex]x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \left(-35\right)}}{2\cdot \:1}[/tex]
[tex]x_{1,\:2}=\frac{-2\pm \:12}{2\cdot \:1}[/tex]
[tex]\mathrm{Separate\:the\:solutions}[/tex]
[tex]x_1=\frac{-2+12}{2\cdot \:1},\:x_2=\frac{-2-12}{2\cdot \:1}[/tex]
[tex]x=\frac{-2+12}{2\cdot \:1}:5[/tex]
[tex]x=\frac{-2-12}{2\cdot1}:-7[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=5[/tex] and [tex]x=-7[/tex]
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