-2b^2-18b^2 Help me. Plz

Mathematics

1 answer:

satela [25.4K]3 years ago

7 0

Answer:

${ \tt{ - {2b}^{2} - 18 {b}^{2} }} \\ = - {20 {b}^{2}}$

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How do you solve x^2 plus 16x plus 63=0?

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$x^2+ 16x + 63=0 \\ \\a=1, \ b=16 , \ c=63 \\ \\ \Delta =b^2-4ac =16^2 -4\cdot1\cdot 63 = 256 -252 = 4 \\ \\x_{1}=\frac{-b-\Delta }{2a}=\frac{-16-\sqrt{4}}{2 }=\frac{-16-2}{2}=-\frac{-18}{2}=-9 \\ \\x_{2}=\frac{-b+\Delta }{2a}=\frac{-16+\sqrt{4}}{2 }=\frac{-16+2}{2}=-\frac{-14}{2}=-7 \\ \\ \\Answer : \ x= -7 \ \ or \ \ x= -9$