Question

Resuelve:
1) 2 x^2 + 7x -4 = 0

2) 6x^2 = 10-11x

3) 20x^2-27x= 14

Answer 1

Voy a resolver estas ecuaciones por el método de factoreso (trinomio de la forma ax²+bx+c)

1)
2x² +7x -4 = 0
2 (2x² +7x -4) = 0
(2x)² +7 (2x) -8 = 0  
(2x +8) (2x -1) = 0
       2
(x +4 ) (2x -1) = 0      → Tenemos dos soluciones
x+4 = 0          2x -1 = 0
x₁ = -4                 2x = 1
                           x₂ = 1/2

2)
6x² = 10 -11x
6x² +11x -10 = 0
6 (6x² +11x -10) = 0
(6x)² +11 (6x) -60 = 0
(6x +15) (6x -4) = 0
       3*2
(2x +5) (3x -2) = 0
2x +5 = 0         3x -2 = 0
2x = -5              3x = 2
x₁ = -5/2               x₂ = 2/3

3)
20x² -27x = 14
20x² -27x -14 = 0
20 (20x² -27x -14) = 0
(20x)² -27 (20x) -280 = 0
(20x -35 ) (20x +8) = 0
         5*4
(4x -7 ) (5x +2) = 0
4x - 7 = 0         5x +2 = 0
4x = 7              5x = -2
x¹ = 7/4               x₂ = -2/5

Answer 2

$1)2x^2+7x-4 = 0 \\ \\ x_{1\ y\ 2}= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ x_{1\ y\ 2}= \frac{-7\pm \sqrt{49-4(2)(-4)}}{2(2)} \\ \\ x_{1\ y\ 2}= \frac{-7\pm \sqrt{49+32}}{4} \\ \\ x_{1\ y\ 2}= \frac{-7\pm \sqrt{81}}{4} \\ \\ x_{1\ y\ 2}= \frac{-7\pm 9}{4} \\ \\ x_1= \frac{-7+9}{4}\to x_1= \frac{2}{4}\to \boxed{x_1= \frac{1}{2} } \\ \\ x_2= \frac{-7-9}{4}\to x_2= \frac{-16}{4}\to \boxed{x_2=-4 }$

$2)6x^2+11x-10 = 0 \\ \\ x_{1\ y\ 2}= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ x_{1\ y\ 2}= \frac{-11\pm \sqrt{121-4(6)(-10)}}{2(6)} \\ \\ x_{1\ y\ 2}= \frac{-11\pm \sqrt{121+240}}{12} \\ \\ x_{1\ y\ 2}= \frac{-11\pm \sqrt{361}}{12} \\ \\ x_{1\ y\ 2}= \frac{-11\pm 19}{12} \\ \\ x_1= \frac{-11+19}{12}\to x_1= \frac{8}{12}\to \boxed{x_1= \frac{2}{3} } \\ \\ x_2= \frac{-11-19}{12}\to x_2= \frac{-30}{12}\to \boxed{x_2= \frac{-5}{2}}$

$3)20x^2-27x-14 = 0 \\ \\ x_{1\ y\ 2}= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ x_{1\ y\ 2}= \frac{27\pm \sqrt{729-4(20)(-14)}}{2(20)} \\ \\ x_{1\ y\ 2}= \frac{27\pm \sqrt{729+1120}}{40} \\ \\ x_{1\ y\ 2}= \frac{27\pm \sqrt{1849}}{40} \\ \\ x_{1\ y\ 2}= \frac{27\pm 43}{40} \\ \\ x_1= \frac{27+43}{40}\to x_1= \frac{70}{40}\to \boxed{x_1= \frac{7}{4} } \\ \\ x_2= \frac{27-43}{40}\to x_2= \frac{-16}{40}\to \boxed{x_2=- \frac{2}{5} }$

Espero que te sirva, salu2!!!!