Answer:
x = 1 or x = -3
Step-by-step explanation:
[tex]ax^{2}[/tex] + bx + c = 0
x = [tex]\frac{-b±\sqrt{b^{2}-4ac } }{2a\\}[/tex]
as [tex]x^{2}[/tex] + 2x - 1 = 2 ⇔ [tex]x^{2}[/tex] + 2x - 3 = 0, its solution is
x = [tex]\frac{-2±\sqrt{2^{2}-4 *1*(-3)} }{2*1}[/tex]
= [tex]\frac{-2±\sqrt{4+12} }{2}[/tex] = [tex]\frac{-2±\sqrt{16} }{2}[/tex] = [tex]\frac{-2±\ 4 }{2}[/tex]
therefore, x = 1 or x = -3
hope this helps! :D
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