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4x(4x+3)=92
We move all terms to the left:
4x(4x+3)-(92)=0
We multiply parentheses
16x^2+12x-92=0
a = 16; b = 12; c = -92;
Δ = b2-4ac
Δ = 122-4·16·(-92)
Δ = 6032
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6032}=\sqrt{16*377}=\sqrt{16}*\sqrt{377}=4\sqrt{377}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{377}}{2*16}=\frac{-12-4\sqrt{377}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{377}}{2*16}=\frac{-12+4\sqrt{377}}{32} $
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