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x^2 + 3x - 18 = 0
Solve using the quadratic formula

Sagot :

[tex]x^2+3x-18=0\\\\a=1;\ b=3;\ c=-18\\\\\Delta=b^2-4ac\to\Delta=3^2-4\cdot1\cdot(-18)=9+72=81\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{81}=9\\\\x_1=\frac{-3-9}{2\cdot1}=\frac{-12}{2}=-6;\ x_2=\frac{-3+9}{2\cdot1}=\frac{6}{2}=3[/tex]
4blueforever
x+6 and x-3 in which the absolute value would be if a positive 3 and a negative 6
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