Question

(2x+5) (3x+4)+1=0 solve this problem









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Answer 1

Step-by-step explanation:

$(2x + 5)(3x + 4) + 1 = 0$

simplifying

${6x}^{2} + 15x + 8x + 20 + 1 = 0$

${6x}^{2} + 23x + 21 = 0$

sum=23 and product=126

$6 {x}^{2} + 18x + 7x + 21 = 0$

$6x(x + 3) + 7(x + 3) = 0$

$(x + 3)(6x + 7) = 0$

$x + 3 = 0 \: \: \: \: \: \: \: \: 6x + 7 = 0$

$x = - 3 \: \: \: \: \: \: \: x = \frac{ - 7}{6}$

Or

By using quadratic equation

where a=6 ,b=23, c=21

$\frac{ - 23 + - \sqrt{ {23}^{2} - 4 \times 6 \times 21 } }{2 \times 6}$

$\frac{ - 23 + - \sqrt{529 - 504} }{12}$

$\frac{ - 23 + - \sqrt{25} }{12}$

by solving this you will get the answer

Mark the answer as brainly plz