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(3x+5)(4x-12)=70
We move all terms to the left:
(3x+5)(4x-12)-(70)=0
We multiply parentheses ..
(+12x^2-36x+20x-60)-70=0
We get rid of parentheses
12x^2-36x+20x-60-70=0
We add all the numbers together, and all the variables
12x^2-16x-130=0
a = 12; b = -16; c = -130;
Δ = b2-4ac
Δ = -162-4·12·(-130)
Δ = 6496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6496}=\sqrt{16*406}=\sqrt{16}*\sqrt{406}=4\sqrt{406}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{406}}{2*12}=\frac{16-4\sqrt{406}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{406}}{2*12}=\frac{16+4\sqrt{406}}{24} $
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