Question

4x^2-9x=5 помогите
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Answer 1

$4x^2-9x=5\4x^2-9x-5=0\D=(-9)^2-4*4*(-5)=81+80=161\D>0 => x_1, x_2\x_1=frac{-(-9)+sqrt{161} }{2*4}=frac{9+sqrt{161}}{8} \x_2=frac{-(-9)-sqrt{161} }{2*4}=frac{9-sqrt{161}}{8}$