Question

4(x+5)(x+6)(x+10)(x+12)-3x²=0

Answer 1

4 (x+5) (x+6) (x+10) (x+12) − 3x^2 = 0

4x^4 + 132x^3 + 1565x^2 + 7920x + 14400 = 0

Factor left side of equation.

(x+8) (2x+15) (2x^2 + 35x + 120) =0

Set factors equal to 0.

x + 8 = 0 or 2x + 15 = 0 or 2x^2 + 35x + 120 = 0

x = −8

or x = − 15/2

or x = −4.6802948509750735

or x = −12.819705149

Answer 2

$4(x+5)(x+6)(x+10)(x+12)-3x^{2}=0 \\ \\ 4(x^{2}+11x+30)(x^{2}+22x+120)-3x^{2}=0 \\ \\ 4(x^{4}+22x^{3}+120x^{2}+11x^{3}+242x^{2}+1320x+30x^{2}+660x+3600)- \\ \\ -3x^{2}=0 \\ \\ 4(x^{4}+33x^{3}+392x^{2}+1980x+3600)-3x^{2}=0 \\ \\ 4x^{4}+132x^{3}+1568x^{2}+7920x+14400-3x^{2}=0$

$4x^{4}+132x^{3}+1565x^{2}+7920x+14400=0 \\ \\ 4x^{4}+32x^{3}+100x^{3}+800x^{2}+765x^{2}+6120x+1880x+14400=0 \\ \\ 4x^{3}(x+8)+100x^{2}(x+8)+765x(x+8)+1880(x+8)=0 \\ \\ (x+8) (4x^{3}+100x^{2}+765x+1880)=0 \\ \\ (x+8)(4x^{3}+30x^{2}+70x^{2}+525x+240x+1880) \\ \\ (x+8)[2x^{2} (2x+15)+35x(2x+15)+120(2x+15)]=0$

$(x+8)(2x+15)(2x^{2}+35x+120)=0 \\ \\ x+8=0 \ \vee \ 2x+15=0 \ \vee \ 2x^{2}+35x+120=0 \\ \\ x=-8 \ \vee \ 2x=-15 \ \vee \ 2x^{2}+35x+120=0 \\ \\ x=-8 \ \vee \ x=- \frac{15}{2} \ \vee \ 2x^{2}+35x+120=0 \\ \\ \Delta=35^{2}-4*2*120=1225-960=265 \\ \\ \sqrt{\Delta} = \sqrt{265} \\ \\ x_{1}= \frac{-35- \sqrt{265} }{4} \\ \\ x_{2}= \frac{-35+ \sqrt{265} }{4}$

$\boxed{x\in \lbrace \frac{-35- \sqrt{265} }{4},-8,- \frac{15}{2}, \frac{-35+ \sqrt{265} }{4} \rbrace }$