Question

y=√-x²+3x+4
-x²+3x+4≥0

Answer 1

Ответ:

$\left[ -1 ;4 \right]$

Объяснение:

$-x^{2} + 3 x + 4 =0\\\\ a=-1 ,\ \ b=3 ,\ \ c=4\\\\ D = b^2 - 4ac = 3^2 - 4\cdot( - 1)\cdot4 = 9 + 16 = 25\\\\\sqrt{D} =\sqrt{25} = 5\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-3-5}{2\cdot(-1)}=\frac{-8 }{-2 }=4\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-3+5}{2\cdot(-1)}=\frac{2}{-2}=-1$

$x\in \left[ -1 ;4 \right]$