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3 years ago

HELP ASAP PLEASE & SHOW UR WORK!! THANK YOU <333

Mathematics

1 answer:

alekssr [168]3 years ago

4 0

Answer:

$\bold{x_1=-\dfrac{3+\sqrt{41}}{2}\approx-4.7\ ,\quad x_2=-\dfrac{3-\sqrt{41}}{2}\approx1.7}$

Step-by-step explanation:

$-2x^2-3x+4=0\\\\a=-2\,,\ b=-3\,,\ c=4\\\\\\x=\dfrac{-b-\sqrt{b^2\pm4ac}}{2a}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(-2)(4)}}{2(-1)}=\dfrac{3\pm\sqrt{9+32}}{-2}\\\\\\x_1=-\dfrac{3+\sqrt{41}}{2}\approx-4.7\ ,\qquad x_2=-\dfrac{3-\sqrt{41}}{2}\approx1.7$

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