Question

4x^2-9x=5 помогите
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Answer 1

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