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${\huge{ax^2+bx+c=mx+t \\ax^2+bx-mx+c-t=0 \\ax^2+(b-m)x^2+c-t = 0 \\x_{1/2}=\dfrac{-(b-m)\pm\sqrt{(b-m)^2-4a(c-t)}}{2a} \\\textit{für }(b-m)^2\geq4a(c-t)}}$
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Tue, 03 Oct 2017 10:43 GMT