Question

me ayudan es para mañana
$- 3x(x - 1) + 4 = 6(x - 1) - 5$
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Answer 1

$Primero desarrollar: -3x(x-1)+4\\-3x\left(x-4\right)+4\\=-3x^2+12x+4\\Ahora desarrolla: 6(x-1)-5\\=6x-6-5\\=6x-11\\-> -3x^2+12x+4=6x-11\\ = -3x^2+12x+15=6x\\= -3x^2+6x+15=0\\Resuelve con la formula general de 2do grado:x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\left(-3\right)\cdot \:15}}{2\left(-3\right)}\\\\\\_{1,\:2}=\frac{-6\pm \:6\sqrt{6}}{2\left(-3\right)}\\x=1-\sqrt{6},\:x=1+\sqrt{6}$

Corona pls

Answer 2

Respuesta:

$estas \: son \: las \: dos \\ \: soluciones \: de \: la \: ecuacion = \\ x=\frac{-\sqrt{21}-1}{2} \\ x=\frac{\sqrt{21}-1}{2}$

Explicación paso a paso:

$-3x(x-1)+4=6(x-1)-5 \\ \\ -3x^{2}+3x+4=6\left(x-1\right)-5 \\ \\ -3x^{2}+3x+4=6x-6-5 \\ \\ -3x^{2}+3x+4=6x-11 \\ \\ -3x^{2}+3x+4-6x=-11 \\ \\ -3x^{2}-3x+4=-11 \\ \\ -3x^{2}-3x+4+11=0 \\ \\ -3x^{2}-3x+15=0 \\ \\ x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-3\right)\times 15}}{2\left(-3\right)} \\ \\ x=\frac{-\left(-3\right)±\sqrt{9-4\left(-3\right)\times 15}}{2\left(-3\right)} \\ \\ x=\frac{-\left(-3\right)±\sqrt{9+12\times 15}}{2\left(-3\right)} \\ \\ x=\frac{-\left(-3\right)±\sqrt{9+180}}{2\left(-3\right)} \\ \\ x=\frac{-\left(-3\right)±\sqrt{189}}{2\left(-3\right)} \\ \\ x=\frac{-\left(-3\right)±3\sqrt{21}}{2\left(-3\right)} \\ \\ x=\frac{3±3\sqrt{21}}{2\left(-3\right)} \\ \\ x=\frac{3±3\sqrt{21}}{-6} \\ \\ x=\frac{3\sqrt{21}+3}{-6} \\ \\ x=\frac{-\sqrt{21}-1}{2} \\ \\ x=\frac{3-3\sqrt{21}}{-6} \\ \\ x=\frac{\sqrt{21}-1}{2} \\ \\ x=\frac{-\sqrt{21}-1}{2} \\ x=\frac{\sqrt{21}-1}{2}$

⭐Saludos!!!

Att:Leo