对一元二次方程一脸懵逼,全还给老师了
Topic source拿笔算下: ax^2 +bx +c = 0 (2ax)^2 +4abx +4ac = 0 (2ax)^2 +4abx +4ac +b^2= b^2 (2ax+ b)^2 = b^2-4ac x1 = (-b+math.sqrz(b^2-4ac))/2a x2 = (-b-math.sqrz(b^2-4ac))/2a
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拿笔算下: ax^2 +bx +c = 0 (2ax)^2 +4abx +4ac = 0 (2ax)^2 +4abx +4ac +b^2= b^2 (2ax+ b)^2 = b^2-4ac x1 = (-b+math.sqrz(b^2-4ac))/2a x2 = (-b-math.sqrz(b^2-4ac))/2a
Denis618
哪位兄dei能指点一二