Solution calculation of the quartic equation

For example, enter a=3, b=6, c=-123, d=-126 and e=1080 Click to solve the quartic equation \(ax^4 + bx^3 + cx^2 + dx + e = 0\)

quartic equation<SPAN class=form> : ax<SUP>4</SUP> + bx<SUP>3</SUP> + cx<SUP>2</SUP> + dx + e = 0<BR/> Note: </SPAN> <span id="reply_content_461683341">Let p and q be the square roots of any two non-zero numbers</span> <p> <SPAN class=form>root.<BR/> p = sqrt(y<SUB>1</SUB>)<BR/> q = sqrt(y<SUB>3</SUB>) <BR/> r = -g / (8pq) <BR/> s = b / (4a) <BR/> x<SUB>1</SUB> = p + q + r - s<BR/> x<SUB>2</SUB> = p - q - r - s<BR/> x<SUB>3</SUB> = -p + q - r - s<BR/> x<SUB>4</SUB> = -p - q + r - s</SPAN> <BR/> <b>Cubic equation:</b><BR/> Discriminate<SPAN class=form>(Δ) = q<SUP>3</SUP> + r<SUP>2</SUP><BR/> q = (3c- b<SUP>2</SUP>)/9<BR/> r = -27d + b(9c-2b<SUP>2</SUP>)<BR/> s = r + math.sqrt(</SPAN>Discriminate<SPAN class=form>)<BR/> t = r - math.sqrt(</SPAN>Discriminate<SPAN class=form>)<BR/> term1 = math.sqrt(3.0)((-t + s)/2)<BR/> r<SUB>13</SUB>= 2 * math.sqrt(q)<BR/> y<SUB>1</SUB>=(- term1 + r<SUB>13</SUB>math.cos(q<SUP>3</SUP>/3) )<BR/> y<SUB>2</SUB>=(- term1 + r<SUB>13</SUB>math.cos(q<SUP>3</SUP>+(2math.PI)/3) )<BR/> y<SUB>3</SUB>=(- term1 + r<SUB>13</SUB>math.cos(q<SUP>3</SUP>+(4math.PI)/3) )</SPAN></p>