chonamasong
Answered

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how to solve it ...

1. 5-2x²-6x
2. (x+3) (x+4)=0

Sagot :

[tex]1.\\5-2x^2-6x=0\\-2x^2-6x+5=0\\\\a=-2;\ b=-6;\ c=5\\\\\Delta=b^2-4ac\to\Delta=(-6)^2-4\cdot(-2)\cdot5=36+40=76\\\\x_1=\frac{-b-\sqrt\Delta}{2a}\to x_1=\frac{6-\sqrt{78}}{2\cdot(-2)}=\frac{6-\sqrt{4\cdot19}}{-4}=\frac{6-2\sqrt{19}}{-4}=\frac{\sqrt{19}-3}{2}\\\\x_2=\frac{-b+\sqrt\Delta}{2a}\to x_2=\frac{3-\sqrt{19}}{2}[/tex]

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[tex]2.\\(x+3)(x+4)=0\iff x+3=0\ \vee\ x+4=0\\\\x=-3\ \vee\ x=-4[/tex]
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